# What Could The Inverse Sine Be?

(D4M) — Here is the video transcript for this video.

00:00
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

00:01
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

00:02
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

00:03
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

00:04
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

00:05
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

00:06
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

00:07
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

00:08
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

00:09
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

00:10
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

00:11
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

00:12
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

00:13
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

00:14
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

00:15
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

00:16
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

00:17
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

00:18
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

00:19
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

00:20
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

00:21
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

00:22
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

00:23
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

00:24
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

00:25
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

00:26
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

00:27
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

00:28
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

00:29
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

00:30
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