All language subtitles for Precalculus-Course_en

This video introduces functions and their\n

between input numbers, usually the x values\n

that sends each input number to exactly one\n

of as a rule or machine, in which you can\n

as output. So, non mathematical example of\n

function, which takes as input and a person,\n

This function satisfies the condition that\n

get sent to exactly one output person. Because\n

biological mother. So that rule does give\n

and just use the the mother function, which\n

no longer gonna be a function, because there\n

right, they could have a biological mother\n

or any number of situations. So since there's,\n

put it as input, and then you'd get like more\n

this rule of functions, that would not be\n

with functions that are described with equations,\n

can have the function y equals x squared plus\n

x squared plus one. Here, f of x is function\n

of y. Notice that this notation is not representing\n

x, instead, we're going to be putting in a\n

of f of x or y. For example, if we want to\n

input for x either in this equation, or in\n

plus one, f of two is going to equal five.\n

for x. So that's going to be five squared\n

evaluate a function on a more complicated\n

case, remember, the functions value on any\n

when you plug in that whole expression for\n

quantity a plus three squared plus one, we\n

plus nine plus one, or a squared plus six\n

complex expression, it's important to keep\n

way, you evaluate the function on the whole\n

to write f of a plus three equals a plus three\n

because that would imply we were just squaring\n

plus three. Sometimes the function is described\n

example, this graph is supposed to represent\n

represent functions. For example, the graph\n

That's because the graph of a circle violates\n

line, and I'll intersect the graph in more\n

the vertical line test, any vertical line\n

that means is a function because every x value\n

to it. Let's evaluate gf two, note that two\n

find the corresponding y value. So I look\n

the graph. If that has that x value, that's\n

of that point looks like three, and therefore\n

the same thing to evaluate g of five, I run\n

it on the x axis. But there's no point on\n

g of five is undefined, or we can say it does\n

y values make sense for a function leads us\n

of a function is all possible x values that\n

the y values that makes sense for the function.\n

five didn't have a corresponding y value for\n

in the domain of our function J. To find the\n

the x values that correspond to points on\n

shadow or projection of the graph onto the\n

like we're hitting all x values, starting\n

So our domain is the x's between negative\n

or we can write this in interval notation\n

To find the range of the function we look\n

on this graph, we can do that by taking the\n

y axis, we seem to be hitting our Y values\n

range is wise between negative five and three\n

with square brackets. If we made a function\n

a graph, one way to find the domain and range\n

possible to find the domain at least more\n

We think about what x values it makes sense\n

need to be excluded, because they make the\n

Specifically, to find the domain of a function,\n

denominator zero. Since we can't divide by\n

make an expression inside a square root sign\n

of a negative number. In fact, we need to\n

any even root negative because we can't take\n

we can take an odd root like a cube root of\n

logarithmic functions, we'll have some additional\n

these two principles should handle all functions\n

examples. For the function in part A, we don't\n

a denominator. So we need to exclude x values\n

we need x squared minus 4x plus three to not\n

4x plus three equal to zero, we can do that\n

or x equals one, so we need to exclude these\n

So if I draw the number line, I can put on\n

of those. And my domain includes everything\n

this means my domain is everything from negative\n

from one to three, together with everything\n

we don't have any denominator to worry about,\n

to exclude any x values that make three minus\n

include all x values for which three minus\n

that inequality gives us three is greater\n

less than or equal to three halves. If I can\n

in interval notation, notice that three halves\n

2x is allowed to be zero, I can take the square\n

not a problem. Finally, let's look at a more\n

square root and his denominator. Now there\n

the denominator to not be equal to zero, and\n

to be greater than or equal to zero. from\n

means that x is not equal to three, and x\n

means that x is less than or equal to three\n

on the number line. x is not equal to three\n

everything except those two dug out points.\n

equal to three halves means we can have three\n

to be in our domain. And to be legit for our\n

to be true. So I'm going to connect these\n

looking for numbers on the number line that\n

that above and purple. So that's everything\n

one because one was a problem for the denominator.\n

that are colored both both colors red and\n

see negative infinity, up to but not including\n

to three halves, and I include three half\n

Also, in this video, we talked about functions,\n

and ranges. This video is a brief introduction\n

on their graphs. In this first example, I've\n

function. Because as the x values go from\n

this can be written more formally, by saying\n

to be x values. So that means that x two is\n

it's bigger than x one. So whenever we have\n

means that f of x two has to be bigger than\n

bigger when the x values are bigger. The second\n

Because here as the x values increased from\n

So again, we can write this more formally\n

x two that's bigger than x one, for example,\n

then the y value f of x two is less than f\n

have a bigger x value, we want to have a smaller\n

gives the graph of a function that's increasing\n

This part of the graph, here the functions\n

from left to right, the y values go down.\n

where the function is decreasing. Now this\n

because as x goes from left to right, the\n

in green, for increasing. And finally, on\n

neither increasing nor decreasing, it's completely\n

the intervals on which the graph is increasing\n

in terms of the x values. It wouldn't make\n

because the y values can be the same for different\n

unique. Describe where this function is decreasing.\n

and negative two, and between four and seven.\n

for less than x less than negative two, for\n

whether I use less than or less than or equal\n

or equal to signs at the endpoints of the\n

existing altogether. And I'll use strict less\n

or stops decreasing in the middle of its domain.\n

where the function is increasing. That's for\n

So I can write that as an inequality as negative\n

describe these intervals in interval notation.\n

negative to open bracket, a cup sign for union,\n

part, and the increasing part is negative\n

just a little bit. Now by putting arrows on\n

are just hard stops, that signifies that the\n

So if I write it like that, then the place\n

the same, it's still going to be from x values\n

decreasing part of the function extends further.\n

the way out to infinity in it, we assume that\n

would write that as for infinity, and similar\n

infinity, negative two, because the function\n

x values go all the way to negative infinity\n

and decreasing parts of functions based on\n

going to be heading from left to right. As\n

that y goes up. Decreasing means that y goes\n

you have to do that in terms of x values.\n

minimums for functions from their graphs.\n

f of x has an absolute maximum at the x value\n

C is as big as it ever gets. We can write\n

that's the y value at x equals C is bigger\n

at some other x value for all x values in\n

c is called the absolute maximum value for\n

given is called an absolute maximum point.\n

absolute maximum value, f of c is the highest\n

absolute maximum point is the point where\n

for a function to have more than one absolute\n

where that highest value is achieved. But\n

maximum value. function f of x has an absolute\n

equals C is as small as it ever gets. We can\n

of c is less than or equal to f of x for all\n

of c is called the absolute minimum value\n

absolute minimum point in the graph of f of\n

graph ever achieves. And the point c f of\n

low y value. For example, this function has\n

eight, and it has an absolute minimum point\n

this function stops here, and has domain from\n

function will have an absolute maximum value\n

for 10. If however, the function keeps going\n

maximum value at all. absolute maximum minimum\n

minimum values. A function f of x has a local\n

x value of c is bigger than any y values nearby,\n

anywhere in the domain. For example, for this\n

it just has to be the biggest y value nearby,\n

X values around C. We can write this by saying\n

x for all x values, and an open interval around\n

value for F. And the point with coordinates\n

function f of x has a local minimum at x equals\n

x, for all x values in an open interval around\n

value. And the point with coordinate c f a\n

might have many local minimum values. In this\n

zero to four, we have a local minimum point\n

is the lowest point anywhere nearby. It also\n

our attention to local maximum points. This\n

to say, because this point here is the highest\n

the highest point anywhere in the domain since\n

So we have a local max point at one two, which\n

use different conventions as far as whether\n

point or not. Some sources do consider it\n

anywhere nearby. But other sources say that\n

technical reason that you can't get an open\n

is not defined in an open interval around\n

I'll just write this as a caution that some\n

max point, because f is not defined. In an\n

maximum and minimum values can also be called\n

pause the video for a moment to mark all local\n

function, as well as all absolute max and\n

here at the point with coordinates approximately\n

it's the highest point anywhere nearby. There's\n

with coordinates to to, since this is the\n

actually existed, it would also be a local\n

But this drilled out circle means that point\n

no local min point here, and no absolute min\n

is considered by some sources to be a local\n

a local max point. Since the function is not\n

our attention to absolute maximum and points\n

function keeps going up and up and up forever.\n

the function just keeps going down and down\n

a lowest value. Now if we're talking about\n

of the local maxima endpoints. So there's\n

And there's a local min value of two this\n

or min values for this function. In this video,\n

points and values. This video is about symmetry\n

is symmetric with respect to the x axis, if\n

axis as the mirror line. This graph here is\n

that the point five three is on this graph.\n

is also on the graph. Similarly, the point\n

two, negative two is on the graph. For any\n

you flip over the x axis will have coordinates\n

the y coordinates. Therefore, we can say that\n

axis. If whenever a point x, y is on the graph,\n

We say that a graph is symmetric with respect\n

with the y axis as the mirror line. This graph\n

is the mirror line. Notice that the point\n

Its mirror image, which is also on the graph\n

same y coordinate, but the opposite x coordinate.\n

x, y, and I want its mirror image with the\n

the point with the same height the same y\n

So I can say that a graph is symmetric with\n

x, y is on the graph, the point negative x,\n

symmetric with respect to the origin if it\n

the origin. This means that if I spin it by\n

with itself. Rotating a graph by 180 degrees\n

So one way to see if a graph has symmetry\n

down and see if it looks exactly the same.\n

I noticed that the point one, negative two\n

the origin by 180 degrees, I get to the point\n

with a point on the graph with coordinates,\n

get the point with coordinates negative two,\n

with a point x, y, and I rotate that 180 degrees\n

negative x, negative y. so we can say that\n

If whenever a point x, y is on the graph,\n

the graph. Please pause the video for a moment\n

with respect to the x axis, the y axis and\n

one type of symmetry. Graph a is symmetric\n

rotate it by 180 degrees, that is you turn\n

It does not have any mirror symmetry, so it's\n

the y axis. Graph B does have mirror symmetry,\n

both symmetric with respect to the x axis\n

It also looks exactly the same upside down.\n

origin. Graph C is symmetric with respect\n

And graph D is symmetric with respect to the\n

words even and odd to describe functions whose\n

is even if its graph is symmetric, with respect\n

with respect to the y axis means that whenever\n

negative x, y is also on the graph. That is,\n

x are the same. Using function notation, I\n

coordinates x, f of x and negative x, f of\n

is given by f of its x value. So we can say\n

X equals f of x, for all x values in its domain,\n

negative x. For example, we can see the function\n

And one way to see this is by looking at its\n

I've drawn. And that is saying that this graph\n

also check that this function is even algebraically.\n

all we have to do is check that F of negative\n

negative x for x into this formula, that's\n

x squared is the same thing as x squared.\n

of x. And the property of being even is satisfied.\n

graph is symmetric with respect to the origin.\n

to the origin means that whenever a point\n

point, negative x, negative y is also on the\n

the graphs y value at negative x are the opposites\n

sign. If we use function notation, for the\n

f of x, and this one is negative x, f of negative\n

at the x value of negative x is just F of\n

see that F of negative X, that y value has\n

y value. A function f of x is odd if F of\n

x in its domain. Let's look at this example.\n

odd by seeing that rotational symmetry around\n

it out algebraically by checking to see if\n

of x. So let's plug in negative x for x in\n

negative x minus one over negative x, that\n

And we can see that that is actually the negative\n

sign, which is negative f of x. So the property\n

about symmetry with respect to the x axis\n

with respect to the origin. functions that\n

even functions and functions that are symmetric\n

There's no word for functions that are symmetric\n

if your graph is symmetric with respect to\n

This video gives the graphs of some commonly\n

The first function is the function y equals\n

this function. If x is zero, y zero, if x\n

to x doesn't have to just be an integer, it\n

the dots we get a straight line through the\n

x squared. If x is zero, y is zero. So we'll\n

y is one, and x is negative one, y is also\n

four and the x value of negative two gives\n

we get a parabola. That is this, this function\n

symmetry across the y axis, the left side\n

of the right side. That happens because when\n

get the exact same y value as when you square\n

The next function y equals x cubed. I'll call\n

x is zero, y is zero. When x is one, y is\n

one, two goes with the point eight way up\n

to give us negative eight. If I connect the\n

this. This function is what's called an odd\n

symmetry occur around the origin. If I rotate\n

words, turn the paper upside down, I'll get\n

has this odd symmetry is because when I cube\n

get n cube the corresponding negative number\n

gives us exactly the negative of the the y\n

Let's look at the next example. Y equals the\n

this function is just x values bigger than\n

square root of a negative number. Let's plug\n

is one square root of one is one, square root\n

get a function that looks like this. The absolute\n

in a few points, x is zero goes with y equals\n

of negative one is one to two is on the graph\n

to I'm ending up getting a V shaped graph.\n

two to the x is what's known as an exponential\n

in the exponent. If I plot a few points, two\n

squared is four, two to the minus one is one\n

fill in a few more points. So let's see, two\n

two gives me 1/4 1/8. Connecting the Dots,\n

might have heard the expression exponential\n

growth, this is function is represents exponential\n

time we increase the x coordinate by one we\n

at a function like y equals three dx, or sometimes\n

about 2.7. These functions look very similar.\n

little more steeply. Now let's look at the\n

when x is zero, but I can plug in x equals\n

one is one, and one over two is one half,\n

first quadrant, but I haven't looked at negative\n

negative one is negative one, whenever negative\n

looking piece in the third quadrant. This\n

an odd function, because it has that 180 degree\n

down, it'll look exactly the same. Finally,\n

Again, it's not defined when x is zero, but\n

see one over one half squared is one over\n

one over two squared is a fourth, it looks\n

just a little bit more extreme rises a little\n

But for negative values of x, something a\n

one over negative two squared is just one\n

that point there, and one over negative one\n

negative values of x is gonna lie in the second\n

an example of an even function because it\n

These are the toolkit functions, and I recommend\n

way, you can draw at least a rough sketch\n

That's all for the graphs of the toolkit functions.\n

the graph of the function changes or transforms\n

rules and examples for transformations of\n

it's helpful if you're already familiar with\n

them toolkit functions, like y equals square\n

absolute value of x and so on. If you're not\n

to watch my video called toolkit functions\n

start by reviewing function notation. If g\n

of x, then we can rewrite these expressions\n

x minus two is the same thing as the square\n

two means we plug in x minus two everywhere\n

as the square root of quantity x minus two.\n

two on the inside of the function, because\n

square root function. Whereas on the first\n

outside of the function. We're doing the square\n

In this third example, g of 3x. We're multiplying\n

evaluate this in terms of square root, we\n

root function, that gives us the square root\n

by three on the outside of the function, this\n

Finally, g of minus x means the square root\n

because we're not used to taking the square\n

if x itself is negative, like negative to\n

or positive two. So we're really be taking\n

case, let me record which of these are inside\n

In this next set of examples, we're using\n

this time, we're starting with an expression\n

it in terms of g of x. So the first example,\n

because I'm taking the square root of x first,\n

g of x plus 17. In the second example, I'm\n

the square root of the whole thing. Since\n

inside of my function. So I write that as\n

this notation means I plug in the entire x\n

gives me exactly square root of x plus 12.\n

root first and then multiplying by negative\n

my function, I can rewrite this as minus 36\n

I take x multiplied by a fourth and then apply\n

as g of 1/4 x by 1/4. x is on the inside of\n

parentheses when I use function notation.\n

of this function, y equals the square root\n

since the square root of four is two, it looks\n

the square root of x minus two, notice that\n

that means we're going to take the square\n

for example, if we start with the x value\n

that's zero, then we subtract two to give\n

one, which under the square root function\n

decreased by two, one minus two is negative\n

under the square root function had a y value\n

or zero, its y value is also decreased by\n

with negative two, one goes with negative\n

graph. Because I subtracted two on the outside\n

by two, which brought my graph down by two\n

root of quantity x minus two. Now we're subtracting\n

we subtract two from x first and then take\n

y value of zero as we had in our blue graph,\n

need our original x to be two. In order to\n

blue graph, we need to be taking the square\n

one, which means that we need to start with\n

our y value of two from our original graph,\n

be two, which means we need to start out by\n

our x minus two is four, so our x should be\n

my corresponding y values of square root of\n

that the graph has moved horizontally to the\n

two units, makes sense because we're subtracting\n

but the minus two on the inside that kind\n

expect it to to move the graph left I might\n

units, but instead, it moves the graph to\n

up by two units, in order to get the right\n

again, the observations we made for these\n

according to the following rules. First of\n

like in our example, y equals the square root\n

values, and result in vertical motions, like\n

you expect. So subtracting two was just down\n

would move us up by two numbers on the inside\n

y equals the square root of quantity x minus\n

in a horizontal motion, these motions go in\n

Remember, the minus two on the inside actually\n

had a plus two on the inside, that would actually\n

in a shift those are called translations,\n

three times a squared of x, that would result\n

I start with the square root of x, and then\n

root of x, that stretches my graph vertically\n

to graph y equals 1/3, times the square root\n

a factor of 1/3. Finally, a negative sign\n

with a graph of y equals the square root of\n

root of negative x, that's going to do a reflection\n

is on the inside of the square root sign.\n

reflection across the y axis. If instead I\n

of x, that negative sign on the outside means\n

the x axis. Pause the video for a moment and\n

these four transformations. In the first example,\n

function. adding or subtracting means a translation\n

the function affects the y value, so that's\n

should take the square root of graph and move\n

like this. In the next example, we're adding\n

But now we're moving horizontally. And so\n

we are going to go to the left by 12 units,\n

And the next example, we're multiplying by\n

on the outside of our function outside our\n

So and multiplication means we're stretching\n

we reflect in the vertical direction, here's\n

before I apply the minus sign, and now the\n

Finally, in this last example, we're multiplying\n

we know that multiplication means stretch\n

it's a horizontal motion, and it does the\n

shrinking by a factor of 1/4, horizontally,\n

a factor of four horizontally. that'll look\n

horizontally by a factor of four looks kind\n

one half. And that's actually borne out by\n

x is the same thing as the square root of\n

same thing as one half times the square root\n

that vertical shrink by a factor of one half\n

of four, at least for this function, the square\n

for transformations of functions, which I'll\n

to changes in the y values or vertical motions.\n

the x values and raise out in horizontal motions,\n

or shifts. multiplying and dividing by numbers\n

putting in a negative sign. correspond corresponds\n

the negative sign is on the inside, and a\n

on the outside. Knowing these basic rules\n

to sketch graphs of much more complicated\n

root of x plus two, by simply considering\n

introduces piecewise functions. A piecewise\n

by two or more different rules that apply\n

this function is you calculate negative x\n

negative 2x plus three if x is greater than\n

negative two, well, negative two is less than\n

f of negative two by plugging in negative\n

four. Next, we want to find f of one. Well,\n

two rules. But because we have a greater than\n

we apply this rule. And so we could plug one\n

negative two times one plus three, which gives\n

to compute f of three, since three is bigger\n

And we plug three into that role. That gives\n

it makes sense to also draw the graph in pieces.\n

x squared, just x squared would be a problem\n

opening down goes through the points negative\n

it looks something like this. Now I've drawn\n

only applies when x is less than one. So I'll\n

than one. And I'll erase the part of the parabola\n

leave an open circle here, when x equals one\n

either. Next, I'm going to draw the second\n

So that's a line with slope negative two,\n

through this point 03. And then it goes over\n

the point one, one, over by one down by two.\nAnd

I can continue and draw this straight line.\n

part where x is greater than or equal to one.\n

one, this part here. This time, I'm going\n

to one, since that point is included on that\n

to think about what happens when x equals\n

where we actually include that point. And\n

include that point, but we kind of have to\n

here. Now the last question asks us if this\n

of continuity that I'm going to use is that\n

whole thing without picking up your pencil.\n

pick up our pencil to get from the jump here,\n

it's got a discontinuity when x is one, which\n

rules. Often piecewise functions will have\n

one role to the next. However, it's possible\n

has no discontinuity if the two pieces happen\n

changed the functions definition slightly,\n

the negative x squared when x is less than\n

if x is bigger than equal to one, then when\n

the same, but the linear piece will be two\n

start right here at one, negative one and\n

That's all for this introduction to piecewise\n

what the function does. So the inverse of\n

the inverse of the function that adds two\n

two from a number. This video introduces inverses\n

function defined by this chart. In other words,\n

four is six, and f of five is one, the inverse\n

1x undoes what f does. Since f takes two to\n

So we write this f superscript. Negative one\n

to five, F inverse takes five to three. And\n

is four. And since f takes five to one, f\n

to fill in the chart. Notice that the chart\n

of values, when y equals f inverse of x are\n

but the x values for f of x correspond to\n

values for f of x correspond to the x values\n

key fact inverse functions reverse the roles\n

y equals f of x in blue. Next, I'll plot the\n

Pause the video for a moment and see what\n

How are the blue points related to the red\n

points and the red points are mirror images\n

key fact is that the graph of y equals f inverse\n

f of x by reflecting over the line y equals\n

the roles of y and x. In the same example,\n

open circle means composition. In other words,\n

compute this from the inside out. So that's\n

and f inverse of three, we see as to similarly,\n

that means we take f of f inverse of three.\n

same thing as computing F of two, which is\n

and compute these other compositions. You\n

take f inverse of f of a number, you get back\n

And similarly, if you take f of f inverse\n

you started with. So in general, f inverse\n

of x is also equal to x. This is the mathematical\n

each other. Let's look at a different example.\n

video for a moment and guess what the inverse\n

the work that F does. You might have guessed\n

root function. We can check that this is true\n

F of the cube root of function, which means\n

back to x. Similarly, If we compute f inverse\n

And we get back to excellence again. So the\n

the cubing function. When we compose the two\n

we started with. It'd be nice to have a more\n

besides guessing and checking. One method\n

of y and x. So if we want to find the inverse\n

x over 3x, we can write it as y equals five\n

x to get x equals five minus y over three\n

multiply both sides by three y. Bring all\n

without y's and then to the right side, factor\n

us f inverse of x as five over 3x plus one.\n

inverse function, f inverse are both rational\n

of each other. And then General, f inverse\n

x. This can be confusing, because when we\n

one or two, but f to the minus one of x means\n

It's natural to ask if all functions have\n

you might encounter. Is there always a function\n

answer is no. See, if you can come up with\n

an inverse function. The word function here\n

between x values and y values, such that for\n

corresponding y value. One example of a function\n

the function f of x equals x squared. To see\n

a function. Note that for the x squared function,\n

both go to number four. So if I had an inverse,\n

negative two, the inverse would not be a function,\n

when you look at a graph of y equals x squared.\n

roles of y and x and flip the graph over the\n

graph over the line y equals x, I get this\n

of a function, because it violates the vertical\n

line test is because the original green function\n

2x values with the same y value. In general,\n

only if the graph of f satisfies the horizontal\n

the graph. And at most one point, pause the\n

four graphs satisfy the horizontal line test.\n

functions would have an inverse function you\n

the horizontal line test. So their functions\n

C and D satisfy the horizontal line test.\n

have inverses. functions that satisfy the\n

One to One functions. Equivalently a function\n

x one and x two. the y value is f of x one\n

Sometimes as I said, f is one to one, if,\n

then x one has to equal x two. As our last\n

where p of x as the square root of x minus\n

same axis as p of x, we get the following\n

equals x. If we try to solve the problem,\n

of x minus two, reverse the roles of y and\n

adding two. Now if we were to graph y equals\n

parabola, it would look like the red graph\n

arm on the left side. But we know that our\n

right arm, we can specify this algebraically\n

bigger than or equal to zero. This corresponds\n

the square root of x, y was only greater than\n

the domain and range of P and P inverse, we\n

x such that x minus two is greater than or\n

root of a negative number. This corresponds\n

two, or an interval notation, the interval\n

see from the graph is our y value is greater\n

zero to infinity. Similarly, based on the\n

values greater than or equal to zero, the\n

of P inverse is Y values greater than or equal\n

If you look closely at these domains and ranges,\n

exactly to the range of P inverse, and the\n

inverse. This makes sense, because inverse\n

domain of f inverse of x is the x values for\n

or the range of F. The range of f inverse\n

to the x values or the domain of f. In this\n

inverse functions. inverse functions, reverse\n

f inverse of x is the graph of y equals f\n

we compose F with F inverse, we get the identity\n

compose f inverse with F, that brings x to\n

other. The function f of x has an inverse\n

f of x satisfies the horizontal line test.\n

of f inverse and the range of f is the domain\n

functions will be important when we study\n

functions. There are different ways of describing\n

how to convert between them. angles are commonly\n

If we're talking about degrees, then a full\n

circle then would be 180 degrees. And if I\n

would be a third of 360 degrees or 120 degrees.\n

all the way around the circle is called two\n

that's the circumference of a circle with\n

of two pi radians, which is pi radians. And\n

that would be a quarter of my two pi radians,\n

between degrees and radians is handed to us\n

radians. Both of these represent half of a\n

to radians. Now, you might be wondering, what\n

convention, if we go in the counterclockwise\n

And if we go in the clockwise direction, that's\n

135 degrees to radians, I could multiply by\n

degrees cancel here, and I end up with negative\n

negative three Pi over four radians. Or if\n

2.3562 radians, up to four decimal places.\n

times pi, it's better to leave it in that\n

I can also go the other direction and convert\n

Now again, I we need to use the fact that\n

it like this, with radians on the top and\n

cancel. So that doesn't work. Instead, so\n

over hi radians, so they got cancel the radians\n

pi times 180 over four pi degrees. Now the\n

degrees. Converting seven radians to degrees\n

over pi radians. If I work out seven times\n

O seven degrees up to two decimal places.\n

is more than 360 degrees, so this is more\n

seven radians is more than two pi radians.\n

Sometimes angles are given in terms of degrees,\n

minutes, and 25 seconds. A minute is defined\n

1/60 of a minute, but since a minute is itself\n

degree, or one over 3600 of a degree. You\n

60 minutes in one degree and they're 3600\n

number of degrees minutes and seconds to a\n

have 32 degrees plus 17 minutes plus 25 seconds.\n

I need to convert the minutes to degrees.\n

If I want the unit's to cancel, I better put\n

to 60 minutes. Similarly, if I want to convert\n

there are 3600 seconds in one degree, I want\n

the seconds on the bottom to cancel with a\n

Okay, now I can write this as 32 degrees plus\n

And because the minutes cancelled, and the\n

my calculator, I get 32.2903 degrees up to\n

direction, and convert a decimal number of\n

I'm starting out with 247 degrees plus a decimal\n

0.3486 degrees, two minutes, well, I know\n

time, I want the minutes on the top and the\n

will cancel. So I just multiply my decimal\n

me copy this down. And now I'm going to take\n

So I know that they're 60 seconds in one minute.\n

by 60, which gives me 54.96 minutes, I can\n

seconds and get 247 degrees, 20 minutes, and\n

of measuring angles is all about unit conversion.\n

use the fact that pi radians corresponds to\n

seconds and degrees, we use the fact that\n

60 seconds corresponds to one minute. If you\n

is called an arc. And a wedge of pi for that\n

how to calculate the length of the arc and\n

and the radius of the circle. the circumference\n

equals two pi times the radius. Let's use\n

pool has a radius of eight meters, find the\n

radians. So this angle from the center of\n

I've drawn the angle is a little bit less\n

would be pi radians, which is 3.141 radians.\n

I know that the total circumference is going\n

want a fraction of the circumference. The\n

the fraction of the circle that the angle\n

The fraction of the circle that I want is\n

total number of radians in the circle, which\n

a conference, which we said was two pi times\n

and I'm left with just 2.5 times eight meters,\n

is related to the angle it spans by considering\n

you get by taking the angle over The total\n

which is two pi r. Since the two PI's cancel\n

arc length is given by theta times r. here,\n

the angle, it's important to note that theta\n

for this to work. That's because when I took\n

the total measure of the circle. So it's important\n

a circle is given by the formula, pi r squared,\n

to find the area of the sector of a circle\n

pi over six radians, we know that the total\n

meters squared, or 100 pi meter squared. But\n

us the area of our sector. So we want to take\n

makes times the area of the circle. Well,\n

angle of pi over six over the total angle\n

that by the 100 pi meter squared, that simplifies\n

pi meter squared as a decimal, that's 26.18\n

general, if you want to find the area of a\n

that the sector spans out times the area of\n

fraction of the circle is given by the angle\n

by the two pi radians in the circle. Since\n

by saying the area is given by theta over\n

of the sector in radians. Again, it's important\n

how we're doing our fraction of our circle\n

of the circle. Notice that if you don't have\n

instead, not a big deal, because we can always\n

using our formula. In this video, we saw that\n

the arc length is given by the formula theta\n

by theta over two times r squared, where r\n

linear and radial speed for a rotating circle,\n

Consider a spinning wheel. The angular speed\n

time, I can write that as angle per time,\n

second if the angle is measured in radians,\n

guess we could do degrees per minute, etc.\n

the rim of the wheel. So that is the distance\n

in a unit of time. I can think of that as\n

something like meters per second for example,\n

we have a ferris wheel with radius 20 meters.\n

we want to find its angular speed and the\n

Ferris wheel is going one revolution every\n

per minute, I want to, I want to find the\n

through in a unit of time. I've already got\n

but I've got to somehow convert revolutions\n

per minute, one revolution, but that on the\n

revolution is going through an angle of two\n

going to be one half times two pi radians\n

linear speed, the speed of a point on the\n

minute, I got to somehow convert radians into\n

so that's all the way around the circle, I'm\n

and distance, so that would be two pi times\n

two PI's cancel as do the radians, and I'm\n

how we got from angular speed to linear speed\n

call the angular speed, omega looks like a\n

problem, we found the linear speed v, by starting\n

it by the circumference divided by two pi.\n

the whole circumference as it goes through\n

circle. So I'll write that down here, circumference\n

is given by the formula two pi r, and the\n

problem, that shows that the linear speed\n

radius of the circle. In this video, we defined\n

linear speed was radial speed times the radius\n

functions, sine, cosine, tangent, secant,\n

For a right triangle, with sides of length\n

sine of theta as the length of the opposite\n

opposite to our angle theta has measure a\n

measure C. So that would be a oversee for\n

as the length of the adjacent side, over the\n

side adjacent to theta. Of course, the hypothesis\n

as a partner, so we don't think of it as the\n

tangent of theta is the opposite side length\n

be a over b. The pneumonic to remember this\n

Cosine is adjacent over hypotenuse. And tangent\n

a relationship between tangent and sine and\n

to sine of theta over cosine of theta. If\n

some of theta over cosine of theta is given\n

divided by cosine, which is adjacent over\n

by flipping and multiplying, the high partners,\n

adjacent, which is, by definition, tangent\n

that are defined in terms of sine, cosine\n

of theta, by definition, that's one over cosine\n

adjacent over the partners, which is the high\n

triangle is C over B. cosecant of theta is\n

one over the opposite over the hypotenuse,\n

And for this triangle, that's going to be\n

as one over tan theta. So that's going to\n

multiply, I get adjacent over opposite, which\n

values for cotangent cosecant and secant are\n

sine and cosine, respectively. Let's use these\n

six trig functions for the angle theta in\n

That's the opposite of our hypothesis. Well,\n

down here and has measured to the high partners\n

five. Cosine theta is adjacent over hypotenuse,\n

But fortunately, I can find it using Bagaran\n

I know that a squared, I'll call the side\n

leg of the triangle is equal to c squared,\n

a squared plus two squared equals five squared,\n

25. A squared is 21. So A is plus or minus\n

about the length of a side of a triangle,\n

to my computation of cosine theta, I can write\n

21. Over high partners, which is five, tangent\n

that's going to be two over the square root\n

one over cosine theta, so that's going to\n

is five over the square root of 21. The reciprocal\n

over sine theta, that's going to be the reciprocal\n

theta is one over tan theta, so it's going\n

root of 21. Over to finally, we'll do an application.\n

of elevation as the angle from the horizontal\n

100 meters, we want to find out how high it\n

to relate the known quantities this angle,\n

the unknown quantity is the opposite side\n

equals opposite of our hypothesis, then we\n

degrees to our unknown amount y, which is\n

solving for y This gives that y is 100 meters\n

to compute sine of 75 degrees. Be sure Use\n

in the 75. When I do the computation, I get\n

places. Notice that we're ignoring the height\n

the definitions of the trig functions, you\n

that secant is the reciprocal of cosine cosecant\n

of tangent. In this video, I'll use geometry\n

angle, a 45 degree angle and a 60 degree angle.\n

angle is to use a right triangle with a 45\n

has I have hotness of length one. Since all\n

to 180 degrees, and we already have 90 degrees\n

be 45 degrees. So we have an isosceles triangle\n

that side length a. If we want sine of 45\n

then sine is opposite over hypotenuse. So\n

is, I'll be able to compute sine of 45 degrees.\n

length squared plus that side length squared\n

that a squared plus a squared equals one squared.\n

So a squared is one half, and a is plus or\n

talking about the length of sides of triangles,\n

customary to rewrite this as the square root\n

is one over the square root of two, and then\n

the top and the bottom by the square root\n

the numerator, and the square root of two\n

root of two over two. So the side lengths\n

can figure out the sine of 45 degrees, by\n

That's, I'm looking at this angle. So opposite\n

one. So the sine of 45 degrees is the square\n

is adjacent over hypotenuse. That's this side\n

square root of two over to over one again,\n

triangle with hypotony is one, we use this\n

hypotony is five, not quite drawn to scale.\n

the computation with this triangle. This time,\n

tells me b squared plus b squared equals five\n

b squared equals 25 over two, B is going to\n

over two, again, I can just use the positive\n

over the square root of two, which is five\n

the denominator, I get five root two over\n

over her partners, which is five square root\n

to the square root of two over two as before,\n

of 45 degrees is also square root of two over\n

and cosine are based on ratios of sides. And\n

they'll have the same ratios of sides. To\n

use this 30 6090 right triangle with hypotony\n

an angle of 30 here, so a total angle of 60\n

So we have a 60 6060 triangle. That's an equilateral\n

this side length has length one, this side\n

is one, which means this short side of our\n

back to my original triangle, let's use the\n

his longer side x. Pythagorean theorem says\n

squared. So x squared plus a fourth equals\n

or minus the square root of three fourths\n

root of three over the square root of four,\n

Now using our original triangle, again, let's\n

We know that sine of 30 degrees is opposite\n

is one half and the hypothesis is one. So\n

is one half cosine of 30 degrees is adjacent\n

of three over two divided by one. To find\n

we can actually use the same green triangle\n

of 60 degrees instead. So sine of 60 degrees.\n

opposite to this angle is the square root\n

over hypotenuse gives us one half. I'll summarize\n

a 30 degree angle corresponds to pi over six\n

is pi over six. Similarly, 45 degrees corresponds\n

to pi over three radians. I recommend that\n

two over two, and root three over two. And\n

two go together. And root two over two goes\n

hard to reconstruct the triangles, you know\n

so it must have the side links where the same\n

has one side length smaller than the other.\n

larger side must be root three over two since\n

three over two is bigger than one half. Doing\n

angles, the smaller angle must be the 30 degree\n

one. In this video, we computed the sine and\n

45 degrees, and 60 degrees. This video defined\n

unit circle, a unit circle is a circle with\n

cosine and tangent in terms of right triangles.\n

theory, you could draw a right triangle with\n

the sign as the length of the opposite side\n

use this method to try to compute sine of\n

we draw this 120 degree angle, and this right\n

to get a right triangle. So instead, we're\n

of radius one. The figure below illustrates\n

related. If you draw right triangles with\n

angles, then the top vertex sweeps out part\n

in more detail. In this figure, I've drawn\n

hypothesis of the triangle is the radius of\n

right triangle is at the origin. And another\n

of the circle, I'm going to call the coordinates\n

right triangle has length a, the x coordinate,\n

the y coordinate. If I use the right triangle\n

right here is the angle theta, then cosine\n

a over one, or a. Notice that a also represents\n

circle, at angle theta from the x axis, I'll\n

use the right triangle definition, that's\n

which is just B. But B also represents the\n

circle at angle theta. For tangent theta,\n

its opposite over adjacent. So that's B over\n

of the point over the x coordinate of the\n

part of a right triangle, because they're\n

now I'll call this angle here theta, I can\n

to calculate the sine and cosine of theta.\n

this line at angle theta, if I mark that to\n

I'm still going to define as the x coordinate\n

and tangent theta as the ratio of the y coordinate\n

circle definition, we always draw theta starting\n

Let's use this unit circle definition to calculate\n

In our figure, we have a unit circle. And\n

x&y coordinates of this point on the unit\n

that lies at angle fee for the positive x\n

cosine fee is equal to the x coordinate. And\n

two, which works out to negative 0.3639 up\n

method for calculating sine cosine and tangent\n

the positive x axis, you draw the angle theta\n

of the point on the unit circle where that\n

is the x coordinate, sine of theta is the\n

ratio. This video gives three properties of\n

be deduced from the unit circle definition.\n

sine and cosine for an angle theta is that\n

of theta is the y coordinate for the point\n

property is what I call the periodic property.\n

are periodic with period two pi. And what\n

angle plus two pi, you get the same thing\n

when we write this down, we're assuming that\n

measure theta in degrees, the similar statement\n

equal to cosine of theta, we can make the\n

plus two pi is equal to sine of the original\n

If we want to measure the angle in degrees,\n

is equal to sine of theta, we can see why\n

of sine and cosine. This is our angle theta,\n

a full turn around the unit circle to our\n

and theta plus two pi are just two different\n

And since sine and cosine give you the y and\n

they have to have the same value. Similarly,\n

theta minus two pi, the minus two pi means\n

circle clockwise, we still end up in the same\n

two pi, the x coordinate of that position\n

of theta minus two pi is the same thing as\n

we add or subtract multiples of two pi. For\n

the same thing as cosine of theta, this time,\n

and still gotten back to the same place. So\n

the same thing as cosine of pi plus four pi,\n

about the unit circle, pi is halfway around\n

x coordinate of this point right here. Well,\n

so cosine of pi must be negative one. If I\n

well, that's sine of negative 360 degrees,\n

as sine of minus 60 degrees. Thinking about\n

start at the positive x axis and go clockwise\n

And so that's one of the special angles that\n

of negative root three over two. And therefore\n

over two the y coordinate. The next property\n

cosine is an even function, which means that\n

as cosine of theta, while sine is an odd function,\n

the negative of sine of theta. To see why\n

And the angle negative theta. A negative angle\n

direction from the positive x axis. The coordinates\n

sine theta, whereas the coordinates of this\n

negative theta. But by symmetry, these two\n

therefore cosine of theta must equal cosine\n

have the same magnitude, but opposite signs.\n

Therefore, sine of negative theta is the negative\n

of theta isn't even or odd function. Well,\n

by definition, is sine over cosine. Well,\n

negative of sine of theta, whereas cosine\n

We're getting negative sine theta over cosine\n

tan of negative theta is the negative of tan\n

last property on this video is the Pythagorean\n

squared plus sine of theta squared is equal\n

with this shorthand notation, cosine squared\n

But this notation, cosine squared theta just\n

it. This property is called the Pythagorean\n

Theorem. Let me draw a right triangle on the\n

the coordinates of this endpoint here are\n

to be a unit circle, the hypothesis of my\n

my right triangle is just cosine theta, same\n

the height of my triangle is the y coordinate\n

theorem says that this side length squared\n

squared. Since one squared is the same thing\n

But the green property is handy for computing\n

vice versa. And this problem, we're told that\n

is an angle that lies in quadrant three. When\n

means the terminal side of the angle lies\n

of t is to use the fact that cosine squared\n

is cosine of t squared plus negative 2/7 squared\n

of t squared plus 4/49 is equal to one. And\n

4/49, which is 4540 nights. taking the square\n

t is plus or minus the square root of 45 over\n

45 over seven. Now, since we're in the third\n

represents the x coordinate of this point,\n

t is going to be negative square root of 45\n

problem using the Pythagorean theorem for\n

fact that sine of t is negative two sevens\n

think of this information as telling us that\n

opposite side is to and whose high partners\n

Tiger in theorem says us a squared plus two\n

four is 49. So a squared is 45. And a is plus\n

worrying about a triangle, I'm going to use\n

to be adjacent over hypotenuse. So that's\n

Now I go back to thinking about positive and\n

in the third quadrant, my co sign needs to\n

in front. This alternative solution uses many\n

and ultimately gets us the same answer. This\n

the periodic property, the even odd property,\n

about the graphs of sine and cosine. I want\n

y equals sine t, where t is in radians. I'm\n

this being the y axis. One way to do this\n

using my knowledge of special angles on the\n

graph, if I convert them all to decimals.\n

the dots to get a graph of y equals cosine\n

continue the graph for t values less than\n

points. Or I could just use the fact that\n

two pi to the my angle T, I'll be at the same\n

be exactly the same. Therefore, my values\n

on this graph, repeat themselves. For example,\n

about like here, it's cosine is the same as\n

this dot here and repeat it over here. Similarly,\n

I get the same value of cosine is when it's\n

and I can continue repeating all my dots.\n

say pi over three. And so my whole graph will\n

on this side, something like this. Since subtracting\n

the same value of cosine. We can also plot\n

it by repetition. Going forward, I'll usually\n

cosine of x and y equals sine of x. When I\n

to an angle, while y refers to a value of\n

of x and y, compared to when we're talking\n

cosine value, and y refers to the sine value.\n

of sine and cosine. The first thing you might\n

graph of sine are super similar to each other.\n

as just being the graph of sine shifted to\n

of x as the sine function of x plus pi over\n

move the graph horizontally to the left by\n

of sine as being constructed from the graph\n

by pi over two, that means we can write sine\n

two, since subtracting pi over two on the\n

by pi over two. Next, let's look at domain\n

all real numbers. Alright, that is negative\n

from negative one to one. That makes sense,\n

circle. The input values for the domain come\n

angle positive negative as big as you want,\n

circle, the output values for the range, that\n

from the coordinates on the unit circle. And\n

one or any smaller than negative one. So that\n

you can tell from the graph, here's cosine,\n

axis, and so it must be even. Whereas the\n

the origin and must be odd. The absolute maximum\n

absolute minimum value is negative one. We\n

period to describe these two functions. The\n

between the maximum and minimum points. Here\n

is the vertical distance between a maximum\n

of the amplitude as the vertical distance\n

as half the vertical distance between a min\n

and the sine function, the amplitude is one.\n

at regular horizontal intervals, the horizontal\n

the period. For y equals cosine of x, the\n

the horizontal distance between successive\n

troughs, or minimum points. algebraically,\n

cosine of x and sine of x plus two pi equals\n

themselves over an interval of two pi and\n

graphed y equals cosine of x and y equals\n

a midline at y equals zero, an amplitude of\n

of functions are functions that are related\n

stretching and shrinking and shifting. This\n

start by graphing the function y equals three\n

function y equals sine x. So I'll graph that\n

this graph vertically by a factor of three,\n

horizontally by a factor of one half. If instead\n

one, this plus one on the outside shifts everything\n

amplitude and period of our original y equals\n

2x, and are further transformed y equals three\n

midline at y equals zero, an amplitude of\n

function, y equals three times sine of 2x.\n

by a factor of one half. So it changes the\n

times two pi, which just pi. Since the two\n

distances, it doesn't affect the midline,\n

is a vertical distance. But the three on the\n

particular, it affects the amplitude, since\n

a factor of three, the amplitude of one get\n

case, the midline doesn't actually change,\n

is still a y value of zero. Now on the third\n

and added one on the outside, so that shifts\n

having a midline at y equals zero, we now\n

doesn't change though, it's still three, because\n

the distance between the mid mine and the\n

still pi since the period is a horizontal\n

affects vertical things. Next, let's graph\n

two times quantity x minus pi over four. This\n

function we graphed on the previous page,\n

if we give the name f of x to that function,\n

then we can get g of x by taking f of x and\n

for x. In other words, g of x is f of x minus\n

idea for graphing g of x the function we want\n

did that on the previous page. And then we\n

four, because that's what you do when you\n

So here's the graph of y equals three sine\n

stretched vertically by a factor of three,\n

half. Now, to graph the function that I want,\n

four to the right. Notice that since I had\n

just read off the horizontal shift. But if\n

sine 2x minus pi over two, which is algebraically\n

and think that I needed to shift over by pi\n

figuring out what the shift is, we're factoring\n

to graph this function, same as the one we\n

the outside, that minus one would just bring\n

to look at midline amplitude, and period for\n

of x, and our final transformed function,\n

x minus pi over four minus one, our original\n

amplitude of one and period of two pi. For\n

stretches vertically, so it makes the amplitude\n

everything down by one. So it brings the midline,\n

the two on the inside, shrinks everything\n

period becomes one half times two pi, which\n

going on, are transferring function shifts\n

shift is sometimes called the phase shift.\n

three sine 2x minus pi over four minus one,\n

sine 2x minus pi over two minus one. This\n

B x minus c plus d, where b is positive. If\n

function with cosine in it, then we know that\n

That's because the original midline of sine\n

by D, we know that the amplitude is going\n

stretches everything vertically by a factor\n

say the amplitude is the absolute value of\n

then that amounts to a vertical reflection,\n

the period of the original sine or cosine\n

B amounts to a horizontal shrink by a factor\n

stretch by a factor of one over b if b is\n

a period of two pi, and we're multiplying\n

be two Pi over B. The trickiest thing is the\n

like to factor out this be from my equation.\n

minus c plus d, I'm going to write y equals\n

plus d. Similarly, if it's a sine function,\n

c over b plus d, then I can read off the horizontal\n

to the right, if C ever be as positive and\n

this might seem backwards from what you're\n

sign there. So if C over B is positive, we're\n

shifts right, if C of b over b is negative\n

and that's why it shifts it to the left. So\n

y equals 1/3, cosine of one half x plus three\n

equals minus five, an amplitude of 1/3, a\n

is four pi, and a horizontal shift. But a\n

to the left, the horizontal shift is sometimes\n

graphs of sinusoidal functions. This video\n

secant, cotangent and cosecant. To gain an\n

of x, I think it's handy to look at the slope\n

The slope of this line is the rise over the\n

and the run is given by cosine of theta. So\n

theta, which is simply tan of theta. So if\n

of x as being the angle and y as being the\n

the angle is zero, the slope is zero. But\n

the slope gets bigger and bigger heading towards\n

negative pi over two, the slope is getting\n

at exactly pi over two and negative pi over\n

is undefined. Using this information, let's\n

we're thinking of x as the angle and y as\n

of negative pi over two and pi over two. So\n

zero, and then it heads up towards positive\n

towards pi over two with an undefined value\n

negative infinity as the angle heads towards\n

value at negative pi over two, you can also\n

pi over two, we have the same line as for\n

two, and therefore this picture repeats. And\n

period not to pi, like sine and cosine, but\n

if you take a line and rotate it by 180 degrees,\n

therefore has the same value of tangent in\n

x intercepts all right values of x of the\n

two pi, etc, you can write that as pi times\n

or negative whole number or zero. This makes\n

of x over cosine of x. And so you're gonna\n

which is where the numerator is zero, and\n

two pi, and so on. From the graph, you can\n

like negative three pi over two, negative\n

two, these values can be written as pi over\n

this makes sense from the definition of tangent,\n

the denominator is zero, and cosine x is zero,\n

two, three pi over two, and so on, the domain\n

So that's going to be everything except for\n

as x such that x is not equal to pi over two\n

or the y values go all the way from negative\n

mentioned previously, is pi. Since the smallest\n

to graph y equals secant x, I'm going to remember\n

with a graph of cosine, I can take the reciprocal\n

the reciprocal of one is one, the reciprocal\n

have a value at pi over two, negative pi over\n

pi over two. When I take the reciprocal of\n

get numbers just greater than one, but I would\n

close to zero, I'm going to get really big\n

Similarly, on the other side, over here, I\n

so their reciprocals will be negative numbers\n

of negative one is negative one. And similarly\n

negative buckets and upside down buckets as\n

has a period of two pi, which makes sense,\n

a range that goes from negative infinity to\n

That makes sense because the range of cosine\n

taking the reciprocal of those values. The\n

asymptotes. Now the vertical asymptotes are\n

of the form pi over two, three pi over two,\n

where k is an odd integer. So the domain is\n

to pi over 2k. For K and odd integer. The\n

any, because you can't take one over something\n

We've seen the graph of y equals tan x and\n

equals cotangent x. It looks similar to the\n

function instead of an increasing one, and\n

in different places. Finally, this green graph\n

related to the graph of sine x, since cosecant\n

the graph of sine x in between, you can see\n

reciprocal. I encourage you to memorize the\n

figure out the details by thinking how about\n

of x, and sine x. This video is about the\n

cotangent and cosecant. Before I graph this\n

parent function, y equals cosecant of x. Recall\n

So the graph of cosecant of x has a vertical\n

puts the vertical asymptotes at pi, two pi,\n

one where sine is one. So here, and so it\n

my fancy function, I know that the two on\n

the plus one shifts up by one, the PI means\n

pi. And so the period, instead of being two\n

be two pi divided by pi, which is two. Finally,\n

shift. But in order to see by how much we\n

pie first. So if I rewrite my function, as\n

plus one, I can see the horizontal shift is\n

in pieces. First, I'll do the vertical motions.\n

stretch by a factor of two. And now I'll shift\n

motion, instead of a period of two pi, I'm\n

everything is going to be shifted left by\n

asymptotes at zero, pi, two pi and so on,\n

shifted over by one half. So that means the\n

a half, a half, one and a half, and so on.\n

is by noting that cosecant is one over sine,\n

does not exist when a sine pi x plus one half\n

is equal to a multiple of pi. canceling the\n

number minus a half. So those are exactly\n

halves, and so on that we've drawn. So once\n

we can see that our graph fits in between\n

and shifted by a half, so it ends up hugging\n

all the other pieces also get shrunk horizontally\n

like this. In this next problem, we're given\n

the shape of the graph. And the fact that\n

graph. But y equals tan x by itself would\n

graph has a period of need to find the horizontal\n

easiest to measure if I sketch in the vertical\n

of negative three to an x value of one. So\n

graph might be something more like tan of\n

that pi equals four B. And so B is pi over\n

related to Y equals tan of pi over 4x. But\n

4x, then that middle point right here would\n

point is at negative one, one, that means\n

up by one, the left by one, I can accomplish\n

over four factored out the up by one I get\n

as possible, there might also be a vertical\n

for that is by plugging in x equals zero and\n

front I might need to get this y intercept\n

pi over four quantity x plus one plus one,\n

pi over four, zero plus one plus one. And\n

is just one, this equation simplifies to two\n

a has to be one. So in fact, this equation\n

summarize, if we have a function of the form\n

a accomplishes a vertical stretch. The number\n

the number D shifts vertically by D. And finally,\n

to rewrite our equation in factored form factoring\n

shift right by C over B. The same things hold.\n

cotangent, we can say the same things for\n

The only thing to be careful of is that B\n

pi over B, simply because the original period\n

the original period of tan and co tan is just\n

tan, secant cotangent and cosecant look like,\n

to graph more complicated functions. This\n

sine inverse cosine inverse and tan inverse.\n

first on the thin black line. This is a graph\n

of a function can be found by flipping the\n

y equals x. I've drawn the flipped graph with\n

the blue dotted line is not the graph of a\n

line test. So in order to get a function,\n

need to restrict the domain of sine of x.\n

with a thick black line. If I invert that\n

x, I get the piece drawn with a red dotted\n

vertical line test. So it is in fact, a function\n

infinity to infinity are restricted sine x\n

over two. It's, its range is still from negative\n

Because I've taken the biggest possible piece\n

a function. The inverse sine function is often\n

a function reverses the roles of y at x, it\n

sine of x, the inverse function has domain\n

pi over two to pi over two, which seems plausible\n

the work a function. So if the function sine\n

inverse sine, or arc sine takes numbers x,\n

of pi over two is one, arc sine of one is\n

arc sine of x is the angle between negative\n

x. y is equal to arc sine x means that x is\n

angles, y who sine is x, right, they all differ\n

y is between negative pi over two and pi over\n

domain restriction in order to get a well\n

notation for inverse sine. Sometimes it's\n

But this notation can be confusing, so be\n

one of x does not equal one over sine of x.\n

is another word for cosecant of x. But sine\n

arc sine of x, the inverse sine function,\n

function, let's go through the same process\n

with a graph of cosine of x, we flip it over\n

line. But the blue dotted line is not a function.\n

our original cosine of x to just be between\n

the vertical line test. So it's a proper inverse\n

from zero to pi, and range from negative one\n

has domain from negative one to one, and range\n

angles to numbers, arc cosine takes us from\n

of pi over four is the square root of two\n

of two over two is equal to pi over four.\n

and pi, whose cosine is x. In other words,\n

to cosine of y. And y is between zero and\n

answers for an angle y whose cosine is x.\n

cosine inverse. And again, we have to be careful.\n

thing as one over cosine of x. one over cosine\n

the negative 1x means arc cosine On the inverse\n

same. Finally, let's look take a look at inverse\n

and black. These vertical lines aren't really\n

asymptotes. So in order to get the actual\n

x, we take just one piece of the tangent function.\n

we flip that over the line y equals x, we\n

function because it satisfies the vertical\n

possible to pick a different piece of the\n

is, yes, we could do that. And on another\n

on our planet, we use the convention that\n

is kind of a convenient choice since it's\n

two examples, our choice of restricted domain\n

that led to a conveniently defined inverse\n

our restricted tan function has domain from\n

include the endpoints in that interval, because\n

at negative pi over two and pi over two, so\n

tan function is from negative infinity to\n

from negative infinity to infinity, and range\n

Once again, tangent is taking us from angles\n

to angles. For example, tangent of pi over\n

is pi over four. So arc tan of x means the\n

over two whose tangent is x. y is equal to\n

of y. And that y is between negative pi over\n

can also be written as 10 to the minus one\n

the inverse trig function, arc tan of x. And\n

is called cotangent of x. And that's all for\n

functions. sine inverse x, also known as arc\n

arc cosine x, and tan inverse x, also known\n

to solving trig equations. Let's start with\n

zero, I want to find all the solutions in\n

get a general formula for all solutions, not\n

by rewriting this equation to isolate the\n

going to write two cosine x equals negative\n

I'm looking for the angles x between zero\n

Since negative one half is one of the special\n

of the unit circle to see that the angle between\n

three, or four pi over three. My answer needs\n

no other spots on the unit circle whose cosine\n

because we can always take one of these angles\n

want to Find all solutions, I can take these\n

and four pi over three, and simply add multiples\n

three plus two pi, or two pi over three minus\n

so on. A much more efficient way to write\n

pi times k, any integer that is any positive\n

I can write four pi over three plus two pi\n

principal solution for pi over three by adding\n

is my final solution. Next, let's look at\n

I'm going to start out by cleaning things\n

this case is tangent. So let me add tangent\n

x equals the square root of three. And so\n

the square root of three over three looks\n

square to three over two, which is a special\n

that my unit circle will again, help me find\n

that tan x is sine x over cosine x. So I'm\n

zero and two pi with a ratio of sine over\n

three, I actually only need to look in the\n

those are the quadrants where a tangent is\n

angles whose either sine or cosine has a squared\n

see that tan pi over six, which is sine pi\n

me one half over root three over two, that's\n

three, which is one over root three. If I\n

So that value works. If I try tan of pi over\n

not equal to root three over three. So pi\n

work out some the values in the third quadrant\n

four pi over three does not. So my answer\n

six, and seven pi over six. Now if I want\n

interval from zero to two pi, I noticed that\n

and add multiples of two pi to it, because\n

over six plus two pi k, and pi over six, sorry,\n

This is a correct answer. But it's not as\n

over six over here on the unit circle is exactly\n

both of these and adding multiples of two\n

by just taking one of them and adding multiples\n

to say that x equals pi over six plus pi times\n

all the same solutions. Because when k is\n

when k is odd, I'll get this family. For example,\n

is just the original seven pi over six. If\n

a period of pi instead of two pi, it makes\n

write the solutions in the form. In this video,\n

sine, or tangent, or the same thing would\n

circle to find principal solutions, Principal\n

and two pi. And then adding multiples of two\n

solutions. For tangent, we noticed that it\n

and add multiples of pi instead of two pi.\n

that can be solved using the unit circle.\n

that don't involve special values on the unit\n

solve. Let's look at the equation two cosine\n

start by simplifying and isolating cosine.\n

t equals one divided by three gives cosine\n

values on the unit circle, but I can locate\n

of a point on the unit circle. So I'm looking\n

is approximately 1/3. So maybe right here,\n

are angles in the first and fourth quadrants.\n

these angles, I can take arc cosine of both\n

arc cosine of 1/3, which, using my calculator\n

of 1.2310. Up to four decimal places, it's\n

convention, the answers to trig equations\n

specified. It's also important to notice that\n

same thing. Since cosine t equals 1/3, has\n

to a certain interval, and even if we restrict\n

two different solutions. On the other hand,\n

solution. And that solution lies between the\n

the 1.231, a radians must be giving us this\n

zero and pi. And to get the other angle this\n

This angle and this angle are the same size.\n

go all the way to pi minus r 1.2312 radians.\n

that two pi minus 1.2310 radians is equal\n

on our interval from zero to two pi, are given\n

emphasize that we found the second solution\n

Now it's straightforward to find all solutions,\n

and add multiples of two pi. It can be helpful\n

our equation, cosine t equals 1/3. We want\n

of t intersects the horizontal line at y equals\n

to our first solution of 1.2310 radiance.\n

our second solution. Notice that it occurs\n

here and all the other initial action points\n

points shifted over by multiples of two pi.\n

x minus one equals two. I'll start by rewriting\n

equivalent to sine x equals three fourths.\n

on the unit circle, we can still use the unit\n

there should be in the interval from zero\n

of a point on the unit circle, we're looking\n

is three fourths. So that might be three fourths\n

about right here. And about right here, so\n

the arc sine of both sides to find one of\n

to be arc sine of three fourths, which my\n

be this angle here. To get the other angle,\n

you can get by going up four pi minus my previous\n

So that's my second angle. Notice that this\n

the first angle, not to pi minus the first\n

we can find all solutions simply by adding\n

solutions on a graph of sine, we're looking\n

intersects with the horizontal line at y equals\n

these two solutions where the second one,\n

and then we have all these additional solutions\n

and negative multiples of two pi from the\n

10x equals four. If we simply take x to be\n

we'll get a single solution. But thinking\n

that graph with the horizontal line y equals\n

two solutions in the interval from zero to\n

of tangent is pi. So any one solution will\n

makes sense by thinking about the unit circle.\n

of a line at angle theta takes on all positive\n

the third quadrant. So there are going to\n

tangent is for. We've already found this one.\n

pi. This gives us our answer for the interval\n

two pi to get all other solutions. Or if we\n

the first solution plus multiples of pi instead\n

by exactly pi. Taking the first solution,\n

all solutions corresponding to both of these\n

our solutions to cosine t equals m, we need\n

two pi a minus this solution and add multiples\n

to the equation, sine t equals n. We can start\n

this and adding multiples of to pi. To find\n

take t equal tan inverse of P, and tan inverse\n

Or we can just take 10 inverse of P, and add\n

we saw were more simple than some, because\n

function always landed us in the first quadrant,\n

all for this video on solving trig equations\n

identities involving trig functions like sine\n

examples that just involve quadratic functions.\n

I can rewrite it x squared minus 6x minus\n

times x plus one equals zero, set the factors\n

plus one equals zero. And that gives me the\n

one. Next, let's look at this more complicated\n

x by multiplying out the right hand side.\n

side. So that gives me x squared minus 6x.\n

equal to x squared minus 6x. That's true no\n

all values of x satisfy this equation, we\n

numbers. The second equation is called an\n

of the variable. The first equation, on the\n

only holds for some values of x and not all\n

and try to decide which of the following three\n

these equations hold for all values of the\n

test them by plugging in a few values of the\n

first equation is not an identity. It does\n

x equals zero, then sine of two times zero\n

zero. So it does hold when x is zero. However,\n

times pi over two, that's the same thing as\n

of pi over two, that's two times one, or two,\n

does not hold for x equals pi over two. Since\n

it's not an identity. The second equation\n

for this by plugging in numbers. For example,\n

one is the same thing as negative of cosine\n

cosine of pi over six plus pi is the same\n

even if we check a zillion examples, that's\n

holds, we could have just gotten lucky with\n

bit stronger evidence by looking at graphs,\n

pie graph, y equals cosine of theta plus pi.\n

over to the left by pi. On the other hand,\n

that's the graph of cosine theta, reflected\n

same graph. So graphing both sides gives us\n

it holds for all values of theta. Now the\n

proof, which we'll do later in the course,\n

sum of two angles. In the meantime, let's\n

C is an identity. And we could build evidence\n

or by graphing the left side and the right\n

the graphs coincided. But for this example,\n

verification. In particular, I'm going to\n

rewrite things and rewrite things until I\n

first thing I'll rewrite is secant and tangent\n

and cosine. Since secant of x is one over\n

cosine x, I can rewrite this expression as\n

x over cosine x. I can clean up those fractions\n

sine squared x over cosine x. Now, I noticed\n

So I can pull them together as one minus sine\n

rewrite the numerator one minus sine squared\n

that cosine squared x plus sine squared x\n

squared x is equal to cosine squared x just\n

So I can replace my numerator, one minus sine\n

one cosine from the top and from the bottom,\n

is the right hand side that I was trying to\n

and the Pythagorean identity allows me to\n

of x, it's an identity. The best way to prove\n

algebra, and to use other identities, like\n

of the equation, till it looks like the other\n

is not an identity is to plug in numbers that\n

not true. Now, if you're just trying to decide\n

not worried about proving it, then I recommend\n

and right sides to see if those graphs are\n

that holds for all values of the variable.\n

called the Pythagorean identities. The first\n

sine squared theta equals one. The second\n

secant squared theta. And the third one goes\n

squared theta. Let's start by proving that\n

equals one. I'll do this by drawing the unit\n

the definition of sine and cosine, the x&y\n

and sine theta, the high partners of my triangle\n

circle. Now the length of the base of my triangle\n

point. So that's equal to cosine theta. The\n

as the y coordinate of this point. So that's\n

right triangles says this side length squared\n

squared. So by the Pythagorean theorem, we\n

squared equals one squared, I can rewrite\n

theta equals one, since one squared is one,\n

notation for cosine theta squared. That completes\n

at least in the case when the angle theta\n

the angle was in a different quadrant, you\n

holds. But I won't give the details here.\n

squared theta plus one equals secant squared\n

identity, which said that cosine squared theta\n

to divide both sides of this equation by cosine\n

left side by breaking apart the fraction into\n

plus sine squared theta over cosine squared\n

squared theta is just one. And I can rewrite\n

of theta squared. That's because when I square\n

and square the denominator. And sine squared\n

Similarly, for cosine squared theta. Now on\n

this fraction as one over cosine theta squared.\n

I just get the one squared, which is one divided\n

I'm almost done. sine theta over cosine theta\n

over cosine theta is the same thing as secant\n

says one plus tan squared theta equals sequencer\n

the identity that we were looking for. The\n

is very similar. Once again, I'll start with\n

squared theta equals one. And this time, I'll\n

break up the fraction on the left. And now\n

over sine theta squared plus one equals one\n

can be written as cotangent. And one over\n

me the identity that I'm looking for. We've\n

one, we proved using the unit circle and the\n

identities, we proved by using the first identity\n

formulas are formulas for computing the sine\n

of two angles, the sine of a difference of\n

of two angles. Please pause the video for\n

it true that the sine of A plus B is equal\n

not true. And we can see by an example, if\n

equals pi, then the sine of pi over two plus\n

over two, which is negative one. Whereas the\n

equal to one plus zero, which is one, negative\n

does not hold for all values of a and b. There\n

hold. For example, if a is zero and B zero,\n

need more complicated formulas. It turns out\n

A plus B is given by sine of A cosine of B\n

A plus B is given by cosine A, cosine B minus\n

with a song, sine cosine cosine sine cosine\n

back up the video and sing along with me,\n

for the sine of a sum of angles and the cosine\n

to figure out the sine and cosine of a difference\n

of sine of A minus B as sine of A plus negative\n

this works out two, sine cosine plus cosine,\n

is even, I know that cosine of negative B\n

of negative b is negative sine of B. So I\n

minus cosine of A sine of B. Notice that this\n

as the formula for the sum is just that plus\n

same trick for cosine of A minus B, that's\n

A cosine minus b minus sine of A sine of negative\n

gives us cosine A cosine B plus sine A sine\n

is almost exactly like the for the sum, just\n

Now let's use the angle sum formula to find\n

Now, 105 degrees is not a special angle on\n

sum of two special angles, I can write it\n

the sine of 105 degrees is the sine of 60\n

this is sine, cosine, cosine, sine. And I,\n

sine of 60 degrees is root three over two\n

of 60 degrees is one half, and sine of 45\n

to root six plus root two over four. For our\n

W, given the values of cosine v, and cosine\n

the first quadrant. Remember, to compute the\n

the two cosines. That wouldn't even make sense\n

point seven would give something bigger than\n

than one. Instead, we have to use the angle\n

of v plus w equals cosine, cosine, minus sine,\n

v and the cosine of W. So I could just plug\n

of v and the sine of W from the given information.\n

So here, I'm going to draw a right triangle\n

angle W. Since I know that the cosine of V\n

over 10. And I can think of that as adjacent\n

decorate my triangles adjacent side with the\n

since I know that the cosine of W is point\n

seven on this adjacent side and attend on\n

Theorem, lets me compute the length of the\n

the square root of 10 squared minus nine squared,\n

And here I have the square root of 10 squared\n

root of 51. I can now find the sine of V as\n

the square root of 19 over 10. And the sine\n

Because we're assuming v and w are in the\n

need to be positive, so we don't need to Jimmy\n

our answers, we can just leave them as is.\n

So we have that cosine of v plus w is equal\n

square root of 19 over 10, times the square\n

works out to a decimal approximation of 0.3187.\n

formulas and use them to compute some values.\n

please watch my other video. Remember the\n

for computing sine of A plus B, and cosine\n

cosine, sine, cosine, cosine minus sine sine.\n

formulas. There are many great proofs of the\n

with you one of my favorites for those who\n

up here, so we'll know what we're trying to\n

by drawing an angle A and angle B on top of\n

to this middle line. And I'm going to extend\n

making a right triangle. Finally, I'll draw\n

just touches its vertices. My rectangle is\n

And I'm going to choose units of measurement,\n

has length one. Now let's stop for a minute\n

Since the top and the bottom edge of the rectangle\n

is a transversal. This angle up here must\n

Also, this skinny angle here must have the\n

is 180 degrees, minus 90 degrees minus this\n

the measure of the angles in a triangle minus\n

label this skinny angle with a. Next, let's\n

Based on the middle right triangle with high\n

down here must be cosine of B, since adjacent\n

side length here must be sine of B. Since\n

we see that sine of B is the hypotenuse of\n

little side here has measure sign a time sign\n

of this angle has to equal sign a. A similar\n

measure cosine A time sign B. Please pause\n

side length of this right triangle. In this\n

A cosine B cosine A cosine B, sine of A plus\n

have a rectangle here. So the opposite sides\n

of A plus B has to equal sine of A cosine\n

exactly the first angle sum formula. Also,\n

is exactly the difference of this side length\n

sine of A sine B. And that's the second angle\n

geometric proof of the angle some formulas.\n

theta and cosine of two theta. Please pause\n

this equation sine of two theta equals two\n

true means always true for all values of theta,\n

false. This equation is false, because it's\n

to see this is graphically, if I graph y equals\n

sine theta, squished in horizontally by a\n

I graph y equals two sine theta, that's like\n

by a factor of two. These two graphs are not\n

formula for sine of two theta. And that formula\n

theta. It's not hard to see why that formula\n

that sine of A plus B is equal to sine A,\n

sine of two theta, which is sine of theta\n

theta, plus cosine theta sine theta. simply\n

this angle, some formula. Well, sine theta\n

sine theta. So I can rewrite this as twice\n

formula. There's also a formula for cosine\n

theta minus sine squared theta. Again, we\n

this comes from. cosine of A plus B is equal\n

B. So if we want cosine of two theta, that's\n

cosine theta cosine theta, minus sine theta\n

This can be rewritten as cosine squared theta\n

the formula above. Now there are a couple\n

are also popular. One of them is one minus\n

is cosine of two theta is two cosine squared\n

two formulas from the original one using the\n

squared theta plus sine squared theta is one.\n

squared theta. If I plug that into my original\n

in instead of cosine squared, I'm going to\n

have a nother minus sine squared theta. So\n

grade data, which is exactly what I'm looking\n

identity to write sine squared theta as one\n

this equation and copy it below. But this\n

right here. So that gives me cosine of two\n

one minus cosine squared theta. That simplifies\n

distributing the negative sign, and combining\n

for sine of two theta. And I have three versions\n

two theta. Now let's use these formulas in\n

theta. If we know that cosine theta is negative\n

quadrant three, we have a choice of three\n

to choose the second one, because it only\n

And I already know my value for cosine theta.\n

but then I'd have to work out the value of\n

two theta is twice negative one over root\n

tenths minus one or negative eight tenths,\n

the equation two cosine x plus sine of 2x\n

is that one of the trig functions has the\n

has the argument of 2x. So I want to use my\n

I'll copy down the two cosine x, and now sine\n

this point, I see a way to factor my equation,\n

of these two terms. That gives me one plus\n

That means that either two cosine x is equal\n

That simplifies two cosine x equals zero,\n

I see that cosine of x is zero at pi over\n

x is negative one at three pi over two, there's\n

is going to be pi over two plus multiples\n

of two pi. This video proved the double angle\n

cosine theta. and cosine of two theta is cosine\n

also proved two alternate versions of the\n

is about the half angle formulas for computing\n

over two. Cosine of theta over two is either\n

cosine theta over two. To figure out whether\n

about what quadrant, the angle theta over\n

two to be negative, then you're gonna need\n

to be positive, then of course, you want the\n

is always positive. Notice that this formula\n

the square root sign is always going to be\n

be any more negative than negative one, we\n

two, and that's plus or minus the square root\n

that the formulas for both cosine of A half\n

in them, they just differ by the positive\n

root. To see why these formulas hold, let's\n

we know that cosine of 2x is equal to two\n

cosine of 2x as one minus two, sine squared\n

cosine of 2x as cosine squared x minus sine\n

to our discussion right now. So I'm going\n

going to make the substitution theta equals\n

is two cosine squared theta over two minus\n

over two. Now, I'm going to solve for cosine\n

sides. Divide both sides by two \n

gives me that cosine of theta over two is\n

of cosine theta plus one over two, which is\n

a similar process out with the second formula.\n

that x is theta over two, that gives me cosine\n

over two. Now I'll solve for sine of theta\n

theta over two to both sides. Next, I'll subtract\n

both sides by two. And finally, I'll take\n

me sine of theta over two is plus or minus\n

over two, which is the formula I was looking\n

know where they come from, let's use them\n

is four fifths, and theta is between pi over\n

quadrant, we want to find the exact values\n

two. Since theta is between pi over two and\n

four and pi over two. So theta over two is\n

that cosine theta over two and sine theta\n

down my half angle formulas. And I know I\n

the positive version. Now unfortunately, I'm\n

cosine theta, so I can't plug in directly,\n

out what cosine might be. I'll draw a right\n

of theta is four fifths, I'll decorate the\n

That means that this side length is going\n

minus four squared, which is the square root\n

theta is going to be either plus or minus\n

Now since I'm in the angle, the second quadrant\n

to be negative three fifths. Now I can plug\n

over two is going to be the square root of\n

to the square root of two fifths divided by\n

over the square root of five. Sine of theta\n

theta, so that's one minus negative three\n

one plus three fifths, or eight fifths over\n

the square root of five. In this video, we\n

and the sine of A half angle and use them\n

finding the lengths of all the sides and the\n

In this example, we're given the length of\n

we know the measure of this right angle is\n

the third angle labeled capital A, and the\n

b and lowercase C. To find the measure of\n

of the three angles of a triangle add up to\n

plus 90 degrees plus a is equal to 180 degrees.\n

minus 49 degrees, which works out to 41 degrees.\n

a couple of possible options. We could use\n

opposite over adjacent is B over 23. So b\n

to 26.46 units. Alternatively, we could use\n

b since now if we're looking at the angle\n

that's a little bit harder to solve algebraically.\n

which means that B is 23 divided by 1041 degrees\n

26.46. The reason we want to use tan in this\n

tan of say 49 degrees relates the unknown\n

that we know the measure of if we had use\n

49 is B over C and we'd have two unknowns,\n

to find the side length C, we can have a few\n

For example, we could use the cosine of 49\n

which is 23 oversee. Solving for C, we get\n

to 35.06 units. Another option would be to\n

we know 23 squared plus b squared equals c\n

plus 26.46 squared equals c squared, which\n

which works out again to 35.06. To review\n

is equal to 180 degrees. We used facts like\n

and similar facts about sine and cosine. And\n

us to find all the angles and side lengths\n

of one side and the angle of one of the non\n

we don't know any of the angles except for\n

lengths. To find the unknown angle theta,\n

overhype hotness, so that's 10 over 15. Cosine\n

this equation relates our unknown angle to\n

in our equation to solve for. To solve for\n

10/15 which is 0.8411 radians, or 48.19 degrees.\n

use the fact that sine of fee is 10 over 15\n

a little easier just to use the fact that\n

us that fee plus 90 plus 48.19 is equal to\n

we can find x, either using a trig function\n

it using a trig function, we could write down\n

10. To find that using the Tyrion theorem,\n

equals 15 squared. I'll use the Pythagorean\n

root of 15 squared minus 10 squared. That\n

use many of the same ideas as in the previous\n

of the angles is 180. The Pythagorean theorem\n

cosine, we also use the inverse trig functions\n

angle. This video showed how it's possible\n

right triangle. And then measures of all the\n

the measure of one angle and one side or from\n

means finding all the lengths of the sides\n

information. The Law of Cosines is a tool\n

not necessarily right triangles. Recall that\n

triangle, like this one, the length of the\n

sides by the formula c squared equals a squared\n

of cosines. As a generalization of the Pythagorean\n

right triangles. loosely speaking, the Law\n

a squared plus b squared plus a correction\n

on the size of the angle, that's opposite\n

left, the angle that's opposite to side C\n

c squared is exactly equal to a squared plus\n

angle opposite decides C prime is a little\n

the side length C prime should be a little\n

previous triangle that had the same side length\n

should be less than a squared plus b squared,\n

equal a squared plus b squared minus a little\n

C double prime is a little bigger than 90\n

should be a little bit longer than the side\n

size legs. So C double prime squared should\n

can write this as C double prime squared is\n

bit. The Law of Cosines says precisely what\n

that for any triangle with sides, a B and\n

c squared is equal to a squared plus b squared\n

what happens with this equation when angle\n

and greater than 90 degrees. If angle C is\n

C is equal to zero. So we have that c squared\n

ordinary Pythagorean Theorem. If on the other\n

from the unit circle, we know that cosine\n

positive number, and we'll be subtracting\n

just like we saw in the picture. Finally,\n

we can see from the unit circle that cosine\n

C is going to be less than zero. And we're\n

which means we're actually adding a little\n

When labeling a triangle, the convention is\n

lowercase letters for the side lengths. And\n

When we wrote the law of cosines, on the previous\n

plus b squared minus two A B, cosine angle\n

a, which side we call B, and which side we\n

is opposite to the side and between these\n

written a squared is equal to b squared plus\n

done is replace C with A, A with B, and B\n

we could put the B squared on the left side\n

c squared minus two a C, cosine B, we can\n

convenient for the problem at hand. Let's\n

lengths and angles in this triangle. By convention,\n

a and the side opposite angle B, little B.\n

angle between them. So that's an S, a s, side\n

law of cosines, with the unknown side, c squared\n

for the variables on the right side. So let's\n

mode to find that c squared is equal to 44.44.\n

to 6.67. Next, I'll use the law of cosines.\n

of the Law of Cosines that has cosine be right\n

squared on the left side and goes b squared\n

a C, cosine B. I can plug in values for all\n

B. To do that, I'll subtract eight squared\n

I'll divide both sides by negative two times\n

gives me that cosine of B is equal to negative\n

that my angle B must be greater than 90 degrees,\n

take cosine inverse of both sides to get that\n

which works out to 96.73 degrees. The last\n

could use the Law of Cosines again, and work\n

here. But a simpler thing to do is to use\n

equals 180 degrees. In other words, a plus\n

So A is going to be 180 minus 37 minus 96.73,\n

example, we were given two sides, and the\n

we're given three side lengths. So I'll call\n

we need to find all three angles. Although\n

the ideas are the same as in the previous\n

the form of the Law of Cosines That has cosine\n

left side. To find angle A, we need the form\n

the right side and little a on the left side.\n

Law of Cosines that has cosine of B on the\n

side. For each of these three equations, I'll\n

and use inverse cosine to find my angles.\n

point 20 degrees, and 87.79 degrees add up\n

is just around off error. In fact, I could\n

the measure of angle C, and the measure of\n

degrees to get the measure of angle B. In\n

used it to solve some triangles. The Law of\n

Theorem, with the correction factor to account\n

Recall that solving a right triangle means\n

the measures of the angles from partial information.\n

us solve triangles that are not necessarily\n

theorem says that for a right triangle, like\n

plus b squared. I like to think of the law\n

Theorem, two triangles that are not necessarily\n

of Cosines says that c squared is equal to\n

factor where the correction factor depends\n

see, in this first triangle, on the left,\n

So we know that c squared is equal to exactly\n

triangle, the angle opposite to side C prime\n

this side, C prime should be a little shorter\n

C prime squared is equal to a squared plus\n

says the angle opposite to side C double prime\n

side length should be a little bit bigger\n

it as a squared plus b squared plus a little\n

this little correction factor is. It says\n

C and angle capital C are opposite to side\n

squared minus two A B, cosine of angle C.\n

90 degrees, cosine of C is going to be a positive\n

quantity, just like we saw in the picture\n

90 degrees, then cosine of c is negative.\n

actually adding a little bit, and we get a\n

a triangle, the convention is to use lowercase\n

letters for the angles, and to put angle A\n

the law of cosines, on the previous page,\n

squared minus two A B, cosine C. But it doesn't\n

call B, and which side the coffee. All that\n

this side, and between these two sides, so\n

equals b squared plus c squared minus two\n

plus c squared minus two a C, cosine B, we\n

most convenient for the problem at hand. Let's\n

the angles of this triangle, by convention,\n

B, side, little B, and the side opposite angle\n

and the angle between them, we have all the\n

cosines. So I'll plug that in. And use a calculator\n

44.44. taking the square root, I get that\n

To find the angle B, I need to use a version\n

of angle B here, so it'll need to mention\n

form b squared. On the left, I'll get a squared\n

the right, I'll plug in values from my side\n

by subtracting eight squared, and 6.67 squared\n

by negative two times eight times 6.67. using\n

B is equal to negative point 1172. The negative\n

must be greater than 90 degrees, which agrees\n

of both sides to get that b is equal to cosine\n

out to 96.73 degrees. The last thing I have\n

Law of Cosines again, and work it out just\n

method is to just use the fact that the sum\n

be 180 degrees minus 37 degrees minus 96.73\n

In the previous example, we were given two\n

angle, were given instead, three side lengths,\n

there are a lot of computations involved,\n

problem. To find angle capital C, we need\n

of capital C on the right side, to find capital\n

that has cosine of B on the right side, and\n

equals a squared plus c squared minus two\n

a little A squared on the left side. So that\n

For each of these three equations, I'll plug\n

of the angle and use inverse cosine to find\n

degrees, 87.79 degrees, and 32 point 20 degrees\n

just a tiny bit off due to roundoff error.\n

by just finding the measure of angle C, and\n

subtracting their sum from 180 degrees to\n

law of cosines and used it to solve some triangles.\n

Pythagorean Theorem with an adjustment factor\n

triangles. The Law of Sines is a tool that\n

necessarily right triangles. It's especially\n

and it can also be used in this situation\n

not between them. The Law of Sines says that\n

two sides, lowercase a, b, and c respectively.\n

of side A is equal to the sine of B over B,\n

C. Notice that we're taking the sine of the\n

side lengths. Suppose we know that angle A\n

Side B is 20 degrees, we want to solve the\n

sides and the measure of the third angle.\n

the third angle by subtracting from 180 degrees.\n

Law of Sines to find my remaining side lengths.\n

of angle A over side length A is equal to\n

I already know three of these quantities,\n

plug in and solve for the unknown side length\n

then multiply both sides by 20 over the sine\n

to 20 sine 55 oversized 58, which works out\n

same method to find side length See, this\n

see in our equation. And we could use either\n

for the other side. I'll plug in values as\n

In the previous example, we use the Law of\n

because we already knew the angles. In this\n

And we're going to need to solve for some\n

write down the law of sines and start solving\n

angle. But there can be two plausible angles\n

if sine of the angle happens to be one half,\n

or 150 degrees. In some cases, this can lead\n

conditions given When this happens, and there\n

ambiguous case. With that warning, let's figure\n

in my given information into the law of signs,\n

where there's only one unknown to solve for.\n

times sine 40 degrees over seven, which works\n

see that there'll be two possible angles with\n

in the second quadrant. If I take sine inverse\n

in the first quadrant, which is 47.27 degrees.\n

so is this angle here. And so the angle that\n

going 180 minus 47.27 degrees, which works\n

an acute angle, the other one and obtuse angle,\n

the possible triangles in both triangles,\n

B is the same But everything else is different.\n

and side in each case. In the left side, I'll\n

C, by taking 180 degrees minus angle B minus\n

90 degrees 92.73 degrees. Now we can use either\n

find the final side length C, I'll use the\n

of angle C into this part of my equation,\n

degrees times seven over sine 40 degrees,\n

triangle, we can carry out the same steps\n

of Sines. To get side length to see, we finish\n

these given side length, and angle. In this\n

triangle in two different situations. In the\n

side between them. So that's an A s, a triangle,\n

we ended up just using the law of science\n

ambiguity. In the second situation, we had\n

and an angle that was not between them. So\n

did find two possible triangles. That would\n

two possible triangles. Sometimes when you\n

the other angles and side lengths, you'll\n

for example, that's got to be negative. Because\n

degrees by but you do have to be on the lookout\n

a triangle with the SSA information given.\n

of a parabola. Its vertex, focus and directrix.\n

all points that are equidistant the same distance\n

horizontal line y equals negative P. I'm assuming\n

the origin will be among those points as it\n

the line. But the other points on the x axis\n

the point. So if I want the set of points\n

of the line, that's going to be a curve that\n

example, a point out here will be the same\n

line. So our intuition is suggesting that\n

a parabola. Let's confirm this with some algebra.\n

x, y, its distance from the point zero P is\n

squared plus y minus p squared. Its distance\n

given by its difference in Y coordinates.\n

y plus P. Let me set these two quantities\n

both sides to get on the left x minus zero\n

y minus p squared equals y plus p squared.\n

plus y squared minus two p y plus p squared\n

The P squares cancel out as do the Y squared,\n

y after move In this negative two p y to the\n

one over four p x squared. And you might recognize\n

transformed by vertically stretched it stretching\n

for pee. You might recall that this lowest\n

This point here is called its focus. And the\n

that the number of P in this equation represents\n

and also represents the distance between the\n

if you're interested in the focus and the\n

more useful than the equivalent form something\n

distance P is more hidden. Now, so far, we've\n

than zero. Let's look for a moment at what\n

we draw the point zero, P is going to be below\n

P is going to be above the x axis because\n

makes it a positive y value. Our parabola\n

works out the same is just in our equation,\n

number instead of a positive one. Or if we\n

more familiar, our coefficient of x squared\n

signifies that the parabola is pointing down.\n

to talk about distances, we should write absolute\n

things sideways by looking at the equation\n

zero. on the x axis. Here, we're assuming\n

and a vertical line at x equals negative P.\n

a parabola pointing sideways. And we can work\n

for it. Starting with an arbitrary point,\n

going to be given by x minus p squared plus\n

line x equals negative P is going to be given\n

going to be x minus negative p, or just x\n

to each other, square both sides like before,\n

squared equals four p x, which I could also\n

or x equals A y squared, where my a corresponds\n

negative, then I probably pointing the opposite\n

same, it'll just have a negative coefficient\n

number of being hidden inside this variable\n

me the distance between the vertex and the\n

the vertex and the directrix. Now let's find\n

some arbitrary point h k instead of the origin,\n

If the parabola had its vertex at the origin,\n

x squared equals four p y, or I could rewrite\n

So this parabola should just be shifted to\n

of functions. We know we can accomplish this\n

putting a plus k on the outside. In other\n

P times x minus h squared. Or we can write\n

y minus k. So this is the original All with\n

by shifting right and up to have a vertex\n

the same. It's just you stuck in a minus h\n

transform parabola will have its focus, a\n

going to be the point h, k plus p, that's\n

down from the vertex. So that's going to be\n

the same for this other parabola pointing\n

case, the same story holds for the parabola\n

y squared equals four p x gets transformed\n

k squared equals four p x minus h. This time,\n

right of my vertex. So here it would be h\n

over by P also from the vertex. So this would\n

works the same way, when the parabola is opening,\n

P is negative. Let's use this information\n

at two, four, and focus at negative one, four.\n

help me see how the parabola is lined up.\n

focus is at negative one, four. So here's\n

has its focus, that means that this problem\n

like this. The directrix is going to be on\n

as the distance between the vertex and the\n

three to minus negative one is three, the\n

the right, and so that would be the line x\n

left, I know I need to use the form of the\n

minus h, where h k is the vertex, so that's\n

negative. And since the absolute value of\n

and the focus, I know that P must be negative\n

get y minus four squared equals four times\n

that to y minus four squared equals negative\n

special forms of the equation for a parabola,\n

x minus h, and x minus h squared equals four\n

the vertex and PS absolute value gives the\n

and between the vertex and the directrix.\n

if p is bigger than zero, and left if p is\n

are opening up if p is bigger than zero, and\n

about ellipses. Recall that a circle can be\n

from a fixed point is a constant. And he lips\n

the sum of the distances from two fixed points\n

focuses or foe sigh The sum of the distances\n

string, which is held constant as I draw the\n

ellipse in more detail. In these pictures,\n

The red line segment that cuts through the\n

and passes through this low sigh, it's called\n

through the middle of the ellipse in the shorter\n

two points at the tips of the ellipse where\n

are called the vertices. And ellipse could\n

consider ellipses that are elongated either\n

direction. Let's find the equation of an ellipse\n

zero, and whose vertices are at negative a\n

elongated in the horizontal direction. For\n

the distance from x y to the first focus,\n

to be some constant. And in fact, it turns\n

show you why. If you look at this point right\n

its distance from the first focus is going\n

from the second focus is just going to be\n

minus c plus a minus C, you get exactly to\n

the distance from a point x y to the first\n

to be the square root of x minus negative\n

distance from x y to the second focus, the\n

of x minus c squared plus y minus zero squared.\n

a fair amount of algebra, this simplifies\n

squared plus a squared y squared equals a\n

we let b squared equals a squared minus c\n

x squared plus a squared y squared equals\n

by a squared b squared, this gives us the\n

over a squared plus y squared over b squared\n

minus c squared, it follows that b is less\n

minor axis. In other words, this point right\n

is the point zero negative B. To see why this\n

The sum of the side lengths that I've drawn\n

So that equals to a, which means each of these\n

triangle is C. So by the Pythagorean Theorem,\n

c squared. Since b squared is defined as a\n

be B. To summarize, for a number a bigger\n

plus y squared over b squared equals one represents\n

direction, whose major axis terminates in\n

minor access terminates in the points zero\n

points negative c zero, and c zero. And the\n

is a squared minus c squared equivalently.\n

b squared from both sides, we can write See\n

The way I remember this is that the biggest\n

going to be a, since a is half the length\n

the length of the minor axis, and definitely\n

and one of the focuses. Since a is the biggest,\n

between a, b and c, the equation must be b\n

can be rearranged to either of these two equations.\n

than B still, but this time, we're dividing\n

x squared by the smaller number squared? Well,\n

y in my equation. And so we end up with the\n

the points, zero a, and zero negative a, whereas\n

axis going in between negative b zero, and\n

here on the y axis again. So in this case,\n

Our previous ellipses were centered at the\n

of an arbitrary point, HK, well, then we need\n

direction, and K in the vertical direction.\n

a squared plus y squared over b squared equals\n

over a squared plus y minus k squared over\n

B, that's for an ellipse elongated in the\n

elongated in the vertical direction, we'll\n

the roles of x and y. So this becomes x squared\n

equals one for the center at the origin. And\n

over b squared plus y minus k squared over\n

below. With some thought we can label key\n

over by an amount a, since a is half the length\n

H plus a, k was this vertex will be h minus\n

b, and this one, h k minus b, the fo cy are\n

label the points on the ellipse that's elongated\n

of the ellipse drawn below, and then we'll\n

direction, right here. Its top vertex is at\n

at the point for negative nine. Its center\n

is right here, at the point four, negative\n

at the end of the minor axis. So this one\n

at negative one, negative three. Since my\n

an equation of the form x minus h squared\n

a squared equals one with a bigger than be\n

to four negative three, a is half the length\n

distance between this point and this point,\n

and see that a is equal to six. B is half\n

the difference in x coordinates of these two\n

that all into my equation, I get that x minus\n

three squared over six squared equals one,\n

the major axis somewhere above and below the\n

need to use the fact that c squared is equal\n

c squared is six squared minus five squared,\n

root of 11, which is a little bit bigger than\n

of squared of 11, above and below the center,\n

three plus a squared of 11 and four, negative\n

detailed the anatomy of ellipse, including\n

and its fossa. This video is about hyperbolas.\n

of points x y, such that the sum of the distances\n

fo cy is a constant. Well, I hyperbola is\n

of the distances between x y and each of two\n

In this picture on the left, the two blue\n

The two red points are the focuses or foe\n

If I look at the distance from x, y to one\n

focus, the difference of those two distances\n

on the hyperbola that I choose. If I draw\n

is called the transverse axis. In this left\n

And the hyperbola itself reminds me of a person\n

picture, our transverse axis between the two\n

itself kind of reminds me of a person standing\n

to be oriented other ways besides horizontally\n

and vertical orientations in this video. If\n

the blue branches of the hyperbola, you'll\n

the vertices. Notice that the vertices are\n

closest to one another. The point halfway\n

between the two foes side is called the center\n

that I've drawn that form an X are not actually\n

asymptotes. And you can think of them as guidelines\n

gets closer and closer to these asymptotes,\n

the asymptotes later. Let's find the equation\n

c zero, and c zero, and vertices at negative\n

at center at the origin. From the distance\n

have a point x, y on the hyperbola, then if\n

first focus, and the distance to the second\n

We're always taking the positive difference\n

take the distance to the left focus minus\n

if we happen to be on the left branch, then\n

focus minus the shorter distance, the left\n

we should always get the same number. And\n

I'll show you why. The reason is, if I take\n

distance to this focus is exactly c minus\n

to that focus is going to be well let's see,\n

So that different distance will be a plus\n

difference, A plus A plus c minus a minus\n

c plus a, which is two a. Now I'll focus on\n

write a formula for its distance to this point.\n

be x minus negative c squared plus y minus\n

distance. Now I subtract the shorter distance,\n

the distance formula gives me x minus c squared\n

equal to A for any point on that right branch.\n

a little bit. After that a chunk of algebra,\n

this equation to see squared minus a squared\n

equals a squared times c squared minus a squared.\n

is appearing a couple times here, I'm going\n

squared b c squared minus a squared. Now I\n

minus a squared y squared equals a squared\n

b squared, you'll have the equation x squared\n

equals one. In this equation, notice that\n

to a vertex. The quantity C is not directly\n

b by this equation, which can also be rewritten\n

Or if you prefer, a squared equals c squared\n

remember that C, which is the distance from\n

the three quantities A, B, and C. So you might\n

I'm going to draw a box, whose center is at\n

in a from either direction left, right, and\n

and down. It turns out that the corners of\n

lines. In other words, the slope of this asymptote,\n

this asymptote that's slipping downwards.\n

a. So I can actually write the equations of\n

at the origin as y equals B over A x and y\n

work for a standing up hyperbola with its\n

me rearrange things. And we'll see what changes\n

In this case, where we have a vertical transverse\n

switched. So our equation, which we could\n

a squared minus x squared over b squared equals\n

equals a squared plus b squared. And the two\n

minus a squared, and a squared equals c squared\n

the center to the vertex. And c still represents\n

our box will extend up and down by a units\n

the slope of the asymptote, the rise over\n

that sloped up and negative a Overby for the\n

which go through the center at the origin\n

over b x and y equals negative a over bx.\n

a is going with x in one form of the equation\n

but it always goes with the positive term,\n

B goes with the negative term, the term that\n

x squared or the Y squared determines the\n

way. So far, we've been considering only hyperbola\n

for them, if we sent her as an arbitrary point,\n

term and subtract k from the y term. The graphs\n

get to the vertices, we just start at the\n

and down by a depending on which way the hyperbola\n

plus a K, and H minus A k. Whereas over here,\n

a. To get to the focuses, we have to go over\n

equals a squared plus b squared. Finally,\n

as before. But this time, they have to go\n

of 00. Their equations are going to be given\n

y minus k is negative B over A x minus h,\n

thing y minus k equals a over b x minus h,\n

over b, x minus h. We've done all the abstract\n

If we're given this hyperbola, we can read\n

the x term is positive, I'm gonna let A be\n

square root of 25. Since the x term is positive,\n

is going to be oriented standing up like this.\n

from the center to each vertex is given by\n

will have coordinates for negative three,\n

eight, negative three. To find the full side,\n

squared plus b squared, or four plus 2529.\n

ballpark position of my first eye here, and\n

the center is six, so six minus square 29,\n

six plus squared of 29 negative three. Finally,\n

the center with slope given by B over A, which\n

five halves. The equations for those lines\n

equals five halves x minus six, and y plus\n

six. This gives a decent rough sketch of the\n

I could also plot some points by plugging\n

In this video, we gave two equations for hyperbolas.\n

is oriented. A has to do with the distance\n

c, which can be figured out from the equation\n

is going to represent the distance from the\n

help determine the slope of the asymptotes.\n

Polar Coordinates give an alternative way\n

plane. Instead of describing a point in terms\n

Cartesian coordinates of the point. When using\n

point in terms of radius r, and an angle theta.\n

and theta is the angle that radius line makes\n

points given in polar coordinates. So the\n

the negative two thirds pi is the value of\n

that I need to go clockwise from the positive\n

I normally would for a positive angle. So\n

I need to go to this line right here. And\n

go eight lines out from the origin. So my\n

point has a radius of five and an angle of\n

that I go counterclockwise starting at the\n

two pi. And here, I've got an extra pi to\n

I need to go five units out from the origin.\n

I could have also labeled this point with\n

more than one way to assign polar coordinates\n

pi over four, and a radius of negative 12.\n

to the other side of the circle before I plot\n

the point at an angle of pi over four and\n

here, I go to the opposite side of the circle,\n

origin, but 180 degrees or pi radians around\n

labeled this point using a positive radius\n

plus pi, or five pi over four. And in general,\n

r theta means the same point as the point\n

pi just makes us jump around to the opposite\n

and Cartesian coordinates, it's handy to use\n

to r cosine theta, y is equal to r sine theta\n

which means that R is plus or minus the square\n

theta is equal to y divided by x. Let's see\n

a point with coordinates, x, y, and draw lines\n

triangle is y, the length of the base is x.\n

is the measure of this interior angle. From\n

adjacent over high partners. So that's x over\n

theta. Similarly, sine theta is opposite over\n

y is equal to r sine theta. That gives us\n

tells us that x squared plus y squared is\n

third equation. Finally, tangent theta is\n

which is the fourth equation. To convert five,\n

coordinates, we just use the fact that x equals\n

So in this case, x is equal to five times\n

times square root of three over two, and y\n

So that's equal to negative five halves to\n

to polar coordinates, we know that negative\n

we need to use the fact that r squared is\n

is negative one squared plus negative one\n

over x, so that's negative one over negative\n

r and theta that satisfy these equations are\n

root of two, and theta could be Pi over four,\n

of two pi to either of these answers. But\n

to the right point. The point with Cartesian\n

in the third quadrant. But if we use a theta\n

of square root of two, that would get us to\n

use the polar coordinates of square root of\n

negative square root of two, and pi over four.\n

either of these values of theta, and get yet\n

polar coordinates. This video talked about\n

Cartesian coordinates and polar coordinates,\n

video introduces the idea of parametric equations,\n

f of x, we can describe the x coordinates\n

third variable t, usually thought of as time.\n

y as a separate function of t. This is especially\n

satisfy the vertical line test, and therefore\n

of y in terms of x. A Cartesian equation for\n

only. parametric equations for a curve give\n

usually T. The third variable is called the\n

graph the parametric equations given here\n

this by finding x and y coordinates that correspond\n

t is negative two, you can calculate that\n

you five and why, when you plug in negative\n

video for a moment and fill in some additional\n

of t. Your chart should look like this. And\n

dots, we get something like this. It says\n

of negative two. And this point over here\n

think of t as time, we're traversing the curve\n

for this curve, we need to eliminate the variable\n

is to solve for t and one equation, say the\n

x, which means that t is one half minus x\n

for t into the second equation and get y equals\n

which simplifies to the quadratic equation,\n

17 fourths. Let's try some more examples.\n

us draw the familiar graph of a circle of\n

since the equations x equals cosine t and\n

a way of describing the x&y coordinates of\n

t equals zero, our curve lies on the positive\n

pi, we traverse the curve once in the counterclockwise\n

circle is given by the equation x squared\n

the trig identity cosine squared t plus sine\n

for cosine t, and y for sine t. Please pause\n

curve and rewrite it as a Cartesian equation.\n

the graph is again a unit circle. But this\n

we actually traverse the circle twice in the\n

double arrow going clockwise. The Cartesian\n

plus y squared equals one. And so we found\n

graph on the X Y axis. Let's take a look at\n

specified for t here. So let's just assume\n

from negative infinity to infinity, our Y\n

between one and negative one. Our x values\n

the graph of this curve has to lie on the\n

parabola. But a parametrically defined curve\n

that y is given by cosine of t. So y can only\n

we're only getting the portion of the parabola\n

zero to pi, I traverse this parabola one time.\n

back again in the other direction. And as\n

infinitely many times. The Cartesian equation\n

squared with the restriction that y is between\n

where we went from parametric equations to\n

a Cartesian equation and rewrite it as a parametric\n

as a function of x. So an easy way to parameterize.\n

y is equal to the square root of t squared\n

restriction in terms of x just translates\n

the copycat parameterization. Since we've\n

but T just copies, whatever x does. In the\n

to t, then we get 25 t squared plus 36. Y\n

we'd have y squared equals 900 minus 25 t\n

square root of this quantity? This is a very\n

that even a function of t here because of\n

a better way to parameterize this curve. Because\n

is a good candidate for parameterizing using\n

sides of the equation by 900, we get 25x squared\n

to one, which simplifies to x squared over\n

If I rewrite this as x over six squared plus\n

x over six equal to cosine of t, and y over\n

for any value of t, x over six and y over\n

cosine squared plus sine squared equals one.\n

six cosine of t, y equals five sine of t,\n

As a final example, let's describe a general\n

point, x, y on the circle, we know that the\n

of the circle is equal to r. So using the\n

root of x minus h squared plus y minus k squared\n

gives us the equation for the circle in Cartesian\n

has radius five, and has Center at the point,\n

minus negative three, that's x plus three\n

25. One way to find the equation of a general\n

with the unit circle and work our way up.\n

centered at the origin is given by the equation\n

we want a circle of radius r centered around\n

everything by a factor of R. So we multiply\n

the center to be at HK instead of at the origin,\n

and add K to all our Y coordinates. This gives\n

equations. to match the Cartesian equation\n

and parametric equations as x equals five\n

plus 17. In this video, we translated back\n

parametric equations with a special emphasis\n

about the difference quotient, and the average\n

related to the concept of derivative and calculus.\n

whose graph right here, a secant line is a\n

the graph of the function. I'm going to label\n

this point here on the graph is going to have\n

a, the second point will have x value B and\n

for a function on the interval from a to b\n

line between the two points A F of A and B\n

over the run, or the change in y over the\n

coordinates f of b minus F of A over the difference\n

average rate of change. To put this in context,\n

of a tree. And x represents time in years,\n

in height, or the amount the tree grows. And\n

so a time period. So this average rate of\n

time period. For example, if it grows 10 inches\n

two years, or five inches per year would be\n

of growth, let's compute the average rate\n

root of x on the interval from one to four.\n

minus f of one over four minus one, well,\n

a square root of one. So that's going to be\n

calling these two locations on the x axis,\n

first location, just x and the second location,\n

horizontal distance between these two locations\n

to label this point on the graph of y equals\n

a y coordinate of f of x. The second point\n

a y coordinate of f of x plus h. a difference\n

using this x x plus h notation. So a different\n

of a function, f of x on the interval from\n

quotient represents the slope of the secant\n

the points with coordinates x, f of x, and\n

formula for the difference quotient. Remember\n

could be written as f of b minus F of A over\n

on the x axis. But now I'm calling instead\n

plus h. So I can rewrite this average rate\n

x plus h minus x. That simplifies a little\n

x, I can cancel the Xs, and I get the difference\n

x over h. The quantity h on this nominator\n

represents a difference in x values. Let's\n

for this function given first, or write down\n

That's f of x plus h minus f of x over h.\n

I do this by shoving in x plus h, everywhere\n

that's going to give me two times x plus h\n

how I use parentheses here. That's important\n

entire x plus h for x. So the entire x plus\n

so the parentheses are mandatory. Similarly,\n

x plus h is squared as it needs to be. I mean\n

now. I can multiply out the x plus h squared,\n

sign. So if I multiply out, I'm gonna get\n

Now I can distribute the two to get 2x squared\n

minus x minus h plus three, these two terms\n

get 4x H. And I think that's as simple as\n

out F of x plus h minus f of x. So that's\n

of x. Again, I need to put the F of X formula\n

whole thing. I'll distribute the negative.\n

out. So the 2x squared and the minus 2x squared\n

zero, and the three and the minus three add\n

squared minus h. Finally, I'll write out the\n

by H. I can simplify this further, because\n

of the numerator. If I factor out this H,\n

h, these two H's cancel, and I'm left with\n

one. This difference quotient will become\n

difference quotient for smaller and smaller\n

and ending up with an expression that has\n

or slope of the function itself. In this video,\n

B minus A to calculate an average rate of\n

h minus f of x over h to calculate and simplify\n