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you know what the sine function is right well in this episode we ask the

question what could the inverse sine function be let’s do some math [Music]

hi everyone welcome back we’re going to begin with the question uh what is the

arc sine function so in order to get to this question the right way

i want to mention that this fact was given to us in um a previous

episode of a different series one in which we were studying algebra but

basically it’s saying that in uh when you talk about one-to-one functions

uh they have inverses so one to one functions have inverses now there’s a

link below in the description where the episode uh for this right here

uh is is there and so it basically says if a function is one to one then it has

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an inverse function and so i go over the meaning of what is one to one mean and

what are inverses and i establish the connection between those two right there

and there’s also a link in below in the description where we talk about

the horizontal line test and how that gives us a test

to determine if a function is one to one or not

so these are important episodes now these these two episodes here uh talk

about from an algebraic point of view so

there’s no trigonometry involved in that at all and so um it’s just basically

talking about functions in general and so we’re going to apply some of our

previous knowledge to the case of the sine function and we’re going to try to

come up with an inverse sine function so these are two important facts you

might want to check out the episode the two episodes for that for

that series there that’s a different series though you don’t need to go watch

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the whole series to understand this video right here

now in another previous episode of this series here and by the way what

is the series here so this series is called trigonometry is fun step-by-step

tutorials for beginners uh you can find the link for the full

series playlist uh in the description also now in this episode we talked about

graphing sine functions for example sine of 2x or 2 times sine x

and we did all kinds of transformations to it like amplitude horizontal vertical

shift changing the period in the amplitude and

phase shift and you know those type of things so let’s look at a sine graph just

briefly so if you’re not sure how to graph the sine graph and how to do those

transformations you might want to check out those previous episodes but any case

we’re going to sketch the graph of sign right here and we’re going to go up like

that and come back down like that and the period is 2 pi

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and so you know we’ve been graphing sine a lot of times and we’ve gotten pretty

quick at it in this series and so here we have these tick marks

here pi over two pi three pi over two and two pi we have a height here of one

and a uh minimum right here of minus one and so

when we’re looking at the graph of y equals sine x what we can say is that

it does not pass the horizontal line test so we’ll say fails the

horizontal line test and the reason why it fails is because i can sketch a

vertical line only sorry horizontal line

it only takes one horizontal line so let me sketch one for you to show that it

fails so if i sketch that horizontal line right there it crosses twice

if it crosses twice or more it fails if you can find a horizontal line where it

crosses twice or more it fails so y equals sine x certainly fails the

horizontal line test so um and what we said in the previous episode

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is if it passes the horizontal line test it’s one to one and then it has an

inverse function and actually opposite of that is true

is it has an inverse if it has an inverse function then it has to pass the

horizontal line test and then it has to be one to one all those three things are

actually different different ways of saying the same thing

so point of this right here is that sine function does not have an inverse

so the sine function does not have an inverse does not have an inverse function

now that’s okay though we’re going to be um

uh hard-headed and and and make up an inverse sine function anyways

and but the way we’re going to do that is by restricting this right here so

that it doesn’t fail the uh horizontal line test and so actually what i want to

do is sketch the graph of sine here and i want to graph uh i graphed you know

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from 0 to 2 pi but now i want to actually graph

this part right here actually too i’m going to graph this part right here

and then coming through here and so you know so we know what this repeats

over here is i just want to draw that out a little bit more

so this right here is pi and this right here is minus pi and this is pi over 2

and this is minus pi over 2 right here minus pi over 2 and the height is 1

and this is -1 and now this is still the same sine

function i just grabbed a little bit more of it over here

now what i want to do is i want to restrict the domain of this function

right here so this is say y equals sine x

and instead of going on and on repeating repeating here

what we’re going to do is we’re going to restrict the

domain here and i’m just going to look at this part right here in red

so i’m going to stop right there at that high point right there and i’m going to

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fall down right here and get this a minimum of minus one and

get the maximum of one so this graph of this sine function right here in red

well it’s not really a sine function it’s just a piece of the sine function

so how can we write that down without having to make a graph well we can just

say that the domain is so the domain is minus pi over two to pi over two

and the range here is still the same so the range is minus one to one

and it’s this function right here this restricted sine function so we can

just call this the restricted sine function as opposed to the sine function

so this restricted sine function right here is what we’re actually going to be

finding the inverse of so the reason why is because this passes

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the horizontal line test no matter where you draw a horizontal line it’s only

going to cross once or less so it’s going to pass horizontal line test

in other words it’s one to one function and in other

words it has an inverse function so this function right here definitely has an

inverse function so we’re going to make it some notation

and sketch the graph of it and so to do that let’s get rid of that right there

and so now what we’re going to do is we’re going to switch the x’s and y’s so

another episode uh was how to find the inverse function

and uh you won’t really need to watch that full episode to understand what

we’re about to do but there is an episode out there to get more in

to review that kind of uh idea how to find the inverse function

how to find the inverse function it’s part of the same series

uh how to find inverse function that these two videos were from

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so those three videos were in a previous uh series how to find the

inverse function so to find the inverse function what we do is

and you can only find the inverse function if you know it’s one to one so

there’s i’ll show you the process it’s not the

same process we’re going to use in that episode but i just find that episode fun

to watch but anyways what i’m going to do now is instead of having sign go

and i’ll draw a sign up here real quick again so sine goes up and down along the

x-axis right and the period is two pi right

one to minus one it’s just going up and down along the x-axis so now when we

switch the x’s and y’s now we’re going to go back and forth along the y-axis so

i’m going to start going up in the positive so i’m going to come up here in

the positive right if we switch x’s and y’s they’re both still positive so i’m

just like this and then we come back up here like this

and i’m just going to keep repeating this pattern here it’s going to go back

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and forth and and so go back right here and like that

and now along the x-axis we had the period and we plotted all the and so

here we’re gonna do the same this right here is

one and minus one so how do i know that because remember we’re switching the x’s

and y’s so the y’s here the max y was the one and this is a minus one so now

this along the x-axis this is going to be a max of one right here and a minimum

of minus one right there and you know so half so one full period is two pi

so that’s where we go uh to the right and then to the left and then we come

back here to the to the y axis and then we’re going to start going to the right

again so the period here is two pi and this is pi

and we’ll cut this in half so pi over 2 and this will be 3 pi over 2.

and so this is the graph of the sine inverse sine relation so let me put that

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down here inverse sine relation and the reason why i’m calling it a

relation is because it’s not a function as you can see this right here fails the

vertical line test so you know i can find a vertical line

where it crosses twice or more so this fails the vertical line test

all right so there we go so what we can do now is we can take this

part of the restriction right here and we’ll just look at that part over here

because the restricted sign inverse right it has this positive part right

here and it has this negative part right here

so that’s going to correspond to right here and then right here

and it’s going to stop right there um and the

um right so we’re going to switch x’s and y’s so this y is minus 1 so that x

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is minus one and the x is minus pi over two and now the y is minus pi over two

so this is going to go so the domain of this right here

the domain of the restricted sign of the sine inverse function

so let’s get rid of this part right here so we’re going to say the domain is

so the domain is the minus one to one and the range is

so the range is the y-axis so it’s going

to be from pi over 2 right here where it stops all the way to minus pi over 2

so i’ll say minus pi over 2 to pi over 2.

and notice how the domain and range are inverses of each other and that’s what

happens when you switch the x’s and y’s the domain are the x values which now

become the y values and the range are the y values that now become the x

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values so this right here is the arc cosine function um

so this is inverse sine relation so now let’s say inverse sine function

inverse sine function or we’ll use this notation here y equals sine with a minus

1 up here so minus 1 right here is not an exponent what it represents this

inverse function or sometimes we’ll use this notation

right up here the arc sine function arc sine

so either one of these two is certainly fine there

so and in terms of what is the arc sine function um well you know this is how

we’re going to use it right here we have

the domain we have the range we have the sketch of the graph right here which

specifies all the points on this function right here so um you know we can

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use this wording right here so y equals sine inverse this one right here

um if and only if if and only if um the x is equal the y is uh

yeah the sine y because we’re switching the x’s and y’s so sine of the y equals

x let’s start with x x equals sine y so in other words

this right here is made from starting with sine

x this is the sine function and when you switch the x’s and y’s you now have the

sine inverse function so this is the sine inverse function

right here but we don’t usually write it with x equals sine y we’re going to

write it with y equals sine inverse x so

these two right here mean the same thing if and only if all right so um

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if we’re going to look at an example now

let’s look at an example now so let’s um see if you have any questions

any questions for the comment section below the video here and

so what we’re going to do is look at example one now and here we go so

we’re going to use uh before we use a calculator though we’re going to use

special values so let me get to the calculator part here in a minute all

right so let’s do our first example here so let’s say example one

um because we’re going to need to start plugging in and and doing values like

that right so for example one let’s say here we have y equals sine inverse of

square root of three over two this should be an easy one because we

are familiar with um you know the sign in fact what is the

angle right here where we’re equal to square root of three over two do you

remember that right we did an episode uh previously on how to make the unit

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circle with all the special angles right here um so

and then for part b here we’re going to be looking at y equals arc sine of one

half and then for part c here we’re going to look at y equals sine inverse of -2

so how do we find these values right here so here we’re going to need a an x in

x and let’s put it up here x is in in the domain here minus 1 to 1

and the y is in minus pi over 2 to the power 2.

and remember that’s the domain of the sine inverse function

and this is the range so i’ll just put it like that so this is domain

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and this is the range in other words x’s can come from here

and the y’s the outputs will be coming from here

so is this x right here coming from this domain so this is square root 3 over 2

which is approximately what is that 0.86 something it’s in

certainly in the domain right here so we’re going to be asking the question here

sine of y equals square root of three over two

right so this is the same thing this is if and only if

and we’re going to try to answer this because we’re more familiar with the

sine than we are the sine inverse function in other words we did lots of

work already with sine right so this is the input

for the sine inverse function so that means it’s the output for the sine

function so i always like to refresh you know

refresh my memory about how these work right because it’s just inverse

functions that’s that’s all it is this is the input for the sine inverse which

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means it’s the output for the sine so this is the output for the sine

inverse so that means it’s the input for sine

and so now we try to solve this problem right here

and you know we know this is pi over three right

and so the reason why this is pi over three is because it’s in the range right

here so this right here the y has to be in the range right here so this is the

domain and range for sine inverse and so the output that we’re looking for

here y which is the input right here because

remember there’s infinitely many values where the sine is going to be square

root of 3 over 2. you just keep going around this unit circle just keep adding

2 pi right so we could say you know pi over 3 plus 2 pi and then pi

over 3 plus 4 pi but none of those will be in the range

remember we have this restriction which makes it all work and so this right here

is going to be the y right here so we have the sine inverse of

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square root of three over two is just simply equal to pi over three

now what i’d like to emphasize here is that uh you know what are you actually

looking for right so if someone gives you a number and they say oh evaluate

this you know what is the output well you know

what are you looking for is an output well you’re looking for another number

yeah but it’s also you can also think about it as a

angle in radians right so this is asking the question sine of what angle is the

output square root of three over two um i don’t want an equal sign there so

basically what i did was i took the problem i

it back into a sine problem and then i solve the sine problem right here

and so let’s do that again here let’s do that for part b right here

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so we’re going to say that you know arc sine so sine of y equals one-half

and now we’re going to ask the question right here where is sine equal to

one-half now you know keep in mind that this is uh

what quadrant here is for the range right so if you’re an angle in between

minus pi over two and pi over two you’re either in quadrant one or quadrant four

right now we know that sine is positive right

because this is positive one-half so this says quadrant one in quadrant four so

where is sine positive so it’s in quadrant one right because this is half

and so what what angle right there where are we at one half right there

for the uh sign right there so it’s gonna be um minus pi over six

is going to be pi over six so we’re looking at sine inverse of

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one half is pi over six and so it’s going to be a quadrant one

angle where we’re looking for a quadrant one angle there what about sine inverse

of two so now if we translate this problem into a sine problem and ask the

question where is sine of y equal to two and then you’ll say oh sine of y equals

2 well that cannot happen so this right here does not exist so 2 is

not in the domain is not in the domain of sine inverse

and another way you can say that is sine inverse of -2 if you input that in

you could say does not exist so an inverse of 2 plug in 2 there

and doesn’t exist so there’s no output so 2 is not in the domain of sine

inverse right there so good so there’s our first three examples there

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now what if we want to use a calculator all right so let’s look at trying to use

a calculator here so i’m going to give you some values

here and i want you to pull out your calculator and make sure that you get

these values here so for a i’m going to give you sine inverse of 0.8492

now the problem with using a calculator though is it’s not going to give us the

exact value in a lot of cases so for part a here we want to use um

you know you want to make sure you’re in radian mode

and you want to you know use the right keystrokes

and so what we’re going to get out here is approximately 58.1

and that’ll be in degrees and so in radians this will be about 1.0145

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and we can say radians if we want and so for part b here we’re going to look at

y equals arc sine of negative zero 0.2317 and is this possible to do is this in

the domain yes and it’s between -1 and 1.

and so again we want to use radian mode and we’ll get this rating right here

we’ll get these ratings right here or if you want to convert it to degrees

and so this right here will be approximately -13.4 degrees and then in radians

this will be negative 0.2338 radians so double check that you can do

those on your calculator or in some kind of software

there’s some examples right there for you all right so now let’s

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look at these uh properties that sign and sign inverse have

if you didn’t get those you know values there um you know send me a send

me a comment and i’ll try to help you out there but let’s look at the um

inverse function properties now so this is about two functions here

in which we’re using sine x and we’re using another uh function i’ll

just call it g and we’re going to be looking at the

restrictions and we’re going to talk about the composition of them

so f composed of g of x and this will be sine of sine is the f

function here so sine of sine inverse of x and this is only good for when

right so the x hair has to go into sine inverse so for x in

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the domain of sine inverse so for x in minus one to one and g of f of x and

so this is equal to x for x in that domain here for x in minus one to one

right has to be in the domain of sine inverse

and now if we switch the order now if we have sine inverse of sine of x

and that would be equal to x and this is for x in now the domain is for sine

but the output for sine right so this is the domain for sine this is the domain

of the restricted sign and so that’s for x n minus pi over 2 to pi over 2.

so this is the domain of the restricted sign

that’s why we can only say for x in this interval right here

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so actually there’s an episode um verifying inverse functions also verifying

verifying inverse functions and so what happens is if you have two

functions they’re inverses if you plug in an x you get out the x if you plug in

an x you get out an x but this isn’t for all x’s all real

numbers right because sine inverse has a

restriction and sine x has a restriction and the reason why they had the

restrictions is because the original sine function right here that we’re

trying to work with isn’t one to one doesn’t pass the

horizontal line test so this is actually the restricted version right here

all right so um now we can use these right here to

start evaluating these functions right here by putting sines and sine inverses

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into each other so let’s look at our second example here so example two

and for part a we’re going to say what is the sine of sine inverse of one half

and so this is sine sine inverse so this x

needs to be between -1 and 1 and it is and so it makes sense so we’re going to

get an output now what we’re going to say is that this

is just going to be an uh one half and so this is since this one half is

one half is in minus one to one and so that’s all there is to part a if

you give me an x that’s in the domain of sine inverse well sine will undo what

sine inverse just did right so this is some angle right here and sine of that

angle well it’s not just any angle it’s an angle that’s determined by sine

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inverse right so they’re going to undo each other they’re inverse functions

and so now let’s do arc sine of sine of pi over four

so sine inverse of sine of pi over four now is this angle right here pi over

four is that in the domain right so the domain was

minus pi over 2 to pi over 2 and that is so this is just going to be equal to pi

over 4. so since pi over 4 is in minus pi over 2 to pi over 2.

so if you know these restrictions right here for the sine inverse and the sine

then you know you can get these answers very quickly however if something is

outside of these domains here then we’ll have to work with that right here so

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let’s try this example c here sine of sine inverse of sine of 5 pi over 6 here

right so you think oh they’re just inverses of each other right just like

on part b here sine inverse of sine they’re just pi over four well that only

works if you’re in the uh domain here so here what we would try to do is to say

sine inverse and break this sine of 5 5 5 pi over 6

into something else so here i’m going to use reference angles

so before i fill this in here i’m going to go think about

sine of five pi over six so where’s sine five pi over six

so five pi over six isn’t quite six pi over six so it’s a pi over six short

so pi over six short of pi so it’s about right there and so the

reference angle right here is pi over six that’s the reference angle

so there’s the angle of five pi over six

and the reference angle right here is pi over six so this is equal to sine of pi

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over 6 but what about the sine in front of it well this is a quadrant two angle

and sine is positive in quadrant two so this is the positive right here so in

other words sine of 5 pi over 6 is the same thing as sine of pi over 6. and we

could go find that value if we want but it’s unnecessary because what i’m going

to do is i’m going to say sine of 5 pi over 6 is equal to sine of pi over 6

so if we’re going to take the sine inverse of sine 5 pi over 6 it’s going

to be the same thing as taking the sine inverse of sine pi over 6. and now

pi over 6 is in this domain right here so this is the restricted domain for the

sign so this is just five so this is just pi over six and i’ll just say since

pi over six is in the domain is n minus pi over 2 to pi over 2. here we go

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so there’s some nice examples there all right so we’ve just begun our under

our study of the inverse or the arc cosine function

in the next episode right here we’re going to go over more properties

of the sine and verse so let’s get started on that right now