Inverse Cosine Example – From Beginner To Ace

Video Series: Trigonometry is Fun (Step-by-step Tutorials for Beginners)

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

00:00
welcome back everyone we’re going to begin with what is the inverse cosine
relation so let’s start off with the cosine graph that we know and love
so i’m going to sketch the graph of cosine right here and just a rough sketch
and so here’s the period two pi and halfway is going to be pi right here
we have a height of one and we have a minus one here
that’s where it starts to repeat and right here halfway is power two
where it hits zero and halfway between these two is three pi over two
where it also hits zero now as we know um from algebra
that this is not a one one-to-one function that it fails the horizontal
line test and we can see that by drawing a
horizontal line just just takes one to fail there’s one horizontal line right
there it crosses twice or more so this fails the horizontal line test so in

00:01
order to make an inverse cosine function we’re going to need to take a look at
the inverse cosine relation and we’re going to need to do a restriction
so we’re going to start with the restriction of cosine
and we’re going to look at on it just from 0 to pi right here so let me put
that part here in in red right here and so this is the
cosine function and we’re making a restriction to it
so this part here in red is just going to be the cosine of x
uh let’s just call it y equals cosine of x and we’re going to be restricting it
from zero to pi on zero to pi it hit the highest is a one and it goes
through all these right here and hits minus one right here at pi
so this is a restricted cosine function and what we’re gonna do now is we’re
gonna look at the inverse cosine relation and so let’s do that right here

00:02
so this goes up and down along the x-axis and so if we switch the x’s and
y’s then this will go back and forth along the y-axis
so i’ll put the x-axis about right here and we’ll start to sketch our graph
of the regular cosine graph but this time we switch the x’s and y so we did
the inverse so this will be the inverse cosine relation and we’ll have the
axes here and for every point x y over here we’ll now have a point where the x
y’s are switched so for example this one right here is zero one so over here
we’ll have one zero so this is a one on the y’s it went between
minus one and one so over here on the x’s it’ll go bouncing back and forth
between minus one and one and over here along the x-axis we have
these tick marks we have zero pi over 2 pi 3 pi over 2 and 2 pi so
let’s put them over here and all right so we have the 0 and then

00:03
we have the pi over 2 and we have the pi and we have the 3 pi over 2. and then
here’s where it starts to repeat right here at 2 pi
so we have these x tick marks which now turn into y tick marks
and so what we can do is so this is the inverse cosine relation
it’s just going up and down and you can tell that the inverse cosine relation
i’m using the word here relation because this is not a function
you can easily find a vertical line that crosses twice or more it fails the
vertical line test it is not a function but let’s look at the restricted cosine
function where is that over here so that’s the x’s are restricted between
0 and y so now over here it’s the y’s that’ll be
restricted between 0 and pi so to be this portion right here so
if we just take a look at this part in the red right here the restricted cosine
function and we look at that on the restricted cosine relation right

00:04
here inverse relation then this right here corresponds right there
and so this is the inverse cosine relation but in red we have the
y equals cosine inverse of x and here the domain is the restriction here 0 to pi
and the range is the range here is between -1 and 1. so over here
because we switch the x’s and y’s we have the inverse function
so now here we have the domain here and i’m putting it red because it’s the
cosine inverse function that we’re building and this so the domain here is from
minus one to one which is the range over here and the range for the
cosine inverse function right here the range here is from zero to pi
which is the domain over here so you can

00:05
see how the inverse functions the domain and range switch
so this is y equals cosine inverse of x and so there’s our inverse cosine
function in red relation in black and there’s the two differences between them
and so let’s write this up here what is the cosine inverse function so here’s
here’s how we sketch the graph of it here’s how we come up with the actual graph
and so what we can say that we can write this in words now
this uh inverse cosine function right here we can say
so is the function so inverse cosine inverse cosine function and
you know we’re going to use the notation so is the function

00:06
there’s more than one notation for it here so we can say cosine inverse
and we’re going to use a minus 1 looking like an exponent but this is not an
exponent it represents that we switch the x’s and y’s
or we can use the notation arc cosine so these are both notations that we use
mathematical notation but the actual uh word is inverse cosine
function so that’s the actual name in you know the english language but these
are the mathematical notation that we use for it and so the domain here
again just to you know [Music] is zero to pi
and sorry the domain is minus one to one and the range is zero to pi okay
and so what we have here is that cosine inverse is equal to some angle y

00:07
if and only if the uh cosine of the angle y is equal to x
and so this right here is the formal definition right here and this is the
mathematical notation right here so we can call it the arc cosine
function as well inverse cosine function or the arc cosine function
all right so now let’s see this in practice here
so let’s go on to all right so we got the domain and range and let’s go on to
uh plugging in some numbers into this function right here and evaluating it
substituting in numbers here so the first one we’re going to do here is um
let’s look at for a here we’ll look at we just want to
plug in some numbers here and get familiar with it so what is arc cosine
or cosine inverse of square root of three over two what is this right here
right and so we want to get familiar with this because

00:08
what does this represent so this right here is an angle
whenever you see arc cosine or cosine inverse you want to think it’s an angle
it’s an angle and so i’m going to name that angle and you know i’m going to use
a greek letter we often use greek letters for angles but the definition is
that cosine of the angle has to be equal
to this in other words we switch x’s and y’s but it’s restricted also so that
angle is between zero and pi and by the way an angle being between zero and pi
restricts it to being in quadrant one or quadrant two
but this says quadrant uh one because cosine is positive in quadrant one if
this had said a negative square root of three over two then we would be looking
in quadrant two but this is in quadrant one
right so we can draw a diagram here if we if we want to here’s our angle theta
quadrant one angle right there it’s between zero and pi

00:09
and what do we know about cosine theta it’s the uh uh adjacent
over the hypotenuse so we could draw this uh triangle right
here if we wanted to but actually we don’t really need to because this is a a
special value right here for cosine that you might remember
in case you don’t remember that let me just you know help you out here a little
bit what if we have theta and cosine theta and sine theta i’m sure you’ve
seen this before but just in case you haven’t what are the special angles 0
pi over 6 over 4 pi over 3 pi over 2 this just represents 0
30 degrees 45 degrees 60 degrees 90 degrees um and then i often like to put here
sine first but it doesn’t really matter um this starts at zero [Music] and
one way to think of zero is to say square root of zero over two square root

00:10
of one over two square root of two over two
this will be square root of three over two and square root of four over two
but you know that’s a nice way to uh remember it
but actually you simplify right square root of zero over two is just zero
and then square root of one over two is just one half
and then square root of two over two and then square root of three over two and
square root two over two that’s just uh square root of four over two is just one
there and any case cosine goes backwards so
one and then square root of three over two square root two over two one half
zero and so in case you don’t have those memorized there that’s one way to
remember them but anyways we already know the cosine square root of three
over two right here is just pi over six so this angle right here is pi over six
we know that cosine pi over six is equal
to square root of three over two so this angle right here theta is pi over six
or if we want to we can put it up here just so that we can

00:11
remember this right here cosine inverse of square root of three over two
is pi over six in other words this is just an angle whenever you look at this
right here think angle you may not always immediately be able
to remember the angle but it is an angle all right so long story short
uh because that’s a lot of extra work there for that problem right there
uh because this is a known value right here all right so what about um
uh another one another example so let’s say we have b here and let’s say we want
to find cosine inverse of 0. so this says that
cosine of so this is equal to an angle let’s call it theta
cosine inverse of 0 is equal to an angle let’s call it theta and what does that
mean well cosine of the angle theta is zero
and the theta the angle is in quadrant one or quadrant two it’s between zero
and pi so this is enough information right here to uniquely determine the theta

00:12
theta is pi over two and so what we have is cosine inverse of
zero is pi over two that is our result right there
for those of you who’d like to draw boxes all right let’s do one more let’s
do a part c here let’s look at cosine inverse of minus one half
so how do we do this well here we go think of this as an angle so i’m going
to call it theta so cosine inverse of minus one half is theta
and what does cosine inverse mean well it means two things that cosine of the
angle is minus one half and the angle is restricted between 0 and pi
so this gives us enough information to uniquely determine theta which is what
we’re looking for so um you know we’re looking in quadrant one or two
but this is says cosine is negative so you know we could draw a diagram if

00:13
we want but you probably remember that cosine of you know
pi over three is one half so this is what we need here to help us if
we have a minus one half and we’re looking in uh quadrant one or two
you know what is the angle that we’re going to be looking for so we’re looking
for 2 pi over 3 that’s the angle that we’re looking for right there
so why is that well cosine of 2 pi over 3 is um you know minus one half and so
that’s how we’re gonna get this angle right here two pi over three um yeah so
really nothing more else to it than that let’s go on to the next question
and let’s talk about why these are inverses of each other

00:14
how can we use that these are inverses that cosine and cosine inverse
are inverses of each other why would that be useful so
first off we have something called the cancellation properties
and so we have that cosine of cosine inverse of x is equal to in fact x
as long as you stay within the domain of cosine inverse and that the output here
is in the domain of cosine but what is the domain of cosine it’s
all real numbers so we don’t really need to worry about this output is in the
domain of cosine but what we need to know is that this x
is in the domain of cosine inverse so this is going to be for x in
the domain of cosine inverse here and we can say what is cosine inverse of

00:15
cosine of x and so this will also be x but again we don’t need to worry about
this x being into cosine because the domain of cosine is all real numbers
what we need to worry about is the output here has to be
in the input of the cosine inverse now in order to make this work though in
order for them to be inverses remember we had to restrict cosine and we
restricted cosine to be between 0 and pi in order to make this work so we need
this x right here to only be between zero and pi
and so we can have less than or equal to x here
so this x output right here is equal to a uh our cosine right here so that means
we’re looking at the range of cosine inverse and the range is between zero
and pi uh now some people use these with x’s um you know you can do that but

00:16
um it doesn’t really matter if you use x’s or y’s or thetas
what matters is is that these are inverses of each other as long
as you look upon the restriction that we gave for the cosine function
which allowed us to uniquely define the inverse function for
for that restricted cosine function so let’s just
make sure that we understand this here so let’s say cosine inverse of
cosine of two pi over three cosine inverse
of cosine two pi over three so um which that’s the second one right here right
so cosine inverse of cosine that’s exactly what we have here cosine inverse
of cosine and so we’re going to look at this angle right here 2 pi over 3
is that between 0 and pi and the answer is yes you can just look
at you know 2 pi over 3 sitting right here this is
1 pi over 3 and this is 2 pi over 3 and you know if you went all the way to pi

00:17
that would be 3 pi over 3. so 2 pi over 3 is certainly in
quadrant two so we have this satisfied right here so this is just going to be
equal to according to property number two two pi over three that’s all you
need to know you cannot just put any angle here and
assume they’re going to cancel each other out it has to be for the restriction
so to see that let’s look at part b here
so this will be cosine inverse of cosine of 5 pi over 3
and here in this example 5 pi over 3 is not in the um
you know it’s not in the domain here it’s not between 0 and pi so we don’t
have or sorry um yeah we’re looking at property number two again so here we’re
going to be looking at for this x being between 0 and pi and it’s not
right it’s greater than 3 pi over 3 it’s 5 pi over 3 right so we can write this

00:18
as cosine inverse and we can just find this
number right here what is cosine of 5 pi over 3
right so we can just straight up do this
and you know for that right here we just get one half so this would be cosine
inverse of one half and then now we can say okay what is
cosine inverse of one-half and that’s just pi over three
just 60 degrees right there so in this example right here when we um
input this x here 5 pi over 3 because it’s not in the domain there
it’s not going to be the answer is not going to be 5 pi over 3.
so what you have to do is um you know break it down and find the actual value
right there right and so i guess in case you don’t remember how
to do that right so cosine of 5 pi over 3 is cosine of the reference angle right

00:19
so where is 5 pi over 3 right so here’s 3 pi over 3 and then 4 pi over 3 and
then 5 pi over 3 is somewhere down here right so there’s 5 pi over 3 angle
and what’s the reference angle it’s pi over three
so this is the same thing as cosine of pi over three
however you have to worry about plus or minus right
and this is in quadrant four and cosine is positive in this quadrant right here
and so that’s how we’re getting you know the the pi over three or
the one half here because you know cosine is 60 degrees this is one half
right there all right so yeah there’s how to find
that in case you don’t remember that but but uh in this example here it’s you
don’t just think you put in x and you have inverses
and you get out the x because cosine is not a one to one function we we’re
looking at the restriction of cosine and
that’s where that comes from right there all right i hope you follow a lot

00:20
and let’s look at our next example now can we find the other compositions so in
other words these two right here are looking at
cosine and cosine inverse what if we had a tangent here or a sine here and a
cosine inverse can we do something like that so let’s see
and let’s start off with this example right here
we’re looking at tangent of arc cosine of two thirds
and so remember remember this part right here whenever you see arc cosine or
cosine inverse you want to think of an angle and i like to call that angle
theta so let’s just say theta is arc cosine of two thirds
and now let’s write out what it means to be arc cosine so it means two things it
means cosine you switch the x’s and y’s this is the input this is the output so
i’m going to switch inputs and outputs so cosine of the angle now is two-thirds

00:21
and then the theta is between zero and pi now there’s actually three things to
think about here we have to think about what theta is
and how to find it you know that we write these two things down
but also not only worry about where sine is positive and negative but also that
this right here this uh output right here this two thirds has to be between
minus one and one in other words if someone someone came here and said oh two
not two thirds just a two then this would be a two here and this
is impossible because remember the cosine is between minus one and one
right cosine never gets larger than one so there’s no way it could be a two
so someone said find tangent of arc cosine of two
well then your answer would be there isn’t such a number you would never be
able to find that exact value it’s just not defined
so this is two-thirds and that’s smaller than one and larger than minus one

00:22
so this actually works it makes sense so we can continue
now we’re what quadrant are we in because cosine is positive
and this says quadrant one or two so that means we’re in quadrant one so
we can actually draw a diagram if we wanted to here’s our angle theta
so these two things right here both tell me that theta is in quadrant one this
says quadrant one or two and cosine is positive and now we have cosine cosine is
adjacent over hypotenuse so we could write 2 over 3 and
we can find this missing side if we want but do we want right so what is tangent
of our angle we’re looking for tangent of theta right that’s we named theta so
tangent of theta tangent is opposite over adjacent so yeah we need to side
here let’s call it h and then we use pythagorean so h squared plus

00:23
four equals nine so h squared is five h is square root of five now you could
say plus or minus uh square root of five but h is positive
because we’re in quadrant one so that’s important there you want to make sure
and choose the right sign because there’s two uh values here plus or minus
square root of five but because we’re in
quadrant one so there’s a reason why not just because i want to draw it but
because uh i’m looking at that there in any case h is square root of five so now
we can answer our problem tangent of arc cosine of two thirds is what is tangent
tangent is opposite over adjacent so square root of five over two there we go
tangent of arc cosine of two thirds is square root of five over two excellent
and let’s see if we can do another one so what is the
because this time i’m using cosine inverse notation

00:24
so we’re looking at tangent of an angle think of cosine inverse as an angle so
this is theta right here and it’s going to be cosine inverse of minus 5 over 13.
so cosine theta is -5 over 13 and the angle is between 0 and pi
so before i go on i usually like to check that this makes sense right here
because otherwise you can just reply back what does not exist or or undefined
so i is -5 over 13 is that between -1 and 1
and the answer is yes so it makes sense to carry on
and so what quadrant is theta in that is in quadrant 1 or quadrant 2
and this right here says cosine is negative so we’re going to be in quadrant two
so there’s our angle theta we don’t know what theta is

00:25
that’s we’re trying to find but we know that cosine is going to be
so this is negative so it’s going to be negative 5 and hypotenuse is always
positive so minus 5 over 13. so there’s our
diagram and i like to make the diagram because
i’m trying to figure out what this is this is tangent of theta
and tangent of theta is going to be opposite over adjacent so again i’m
going to need this side right here h so i have h squared plus minus 5 squared
is 13 squared so 8 squared is 169 minus 25 144
so h is positive because it’s going in quadrant 2 h is positive going up so h
is 12. so now here we go tangent of our angle
theta what is tangent of theta it’s 12 over minus 5.

00:26
so we got what we need tangent of cosine inverse of minus 5 over 13 is just
the tangent is opposite all right so 12 over minus 5. there we go
this is a lot of fun all right so let’s do another example
this time let’s do a cosecant cosecant inverse let’s see if we can do one more
all right so theta is and then i’m going to show you something
new how do you do all this without numbers right that’s that’s going to be
more fun right now we have theta is cosine inverse of 1 over square root of 5
and so this is cosine theta is one over square root of five
and the angle is between zero and pi so is this less than one yes
this is less than one you can go find the decimal
approximation for it if you want but it doesn’t matter we’re just making sure

00:27
that it’s less than one it makes sense remember cosine can never be greater
than one this makes sense at least so let’s try to figure out what theta is
all right um i call that just a quick sanity check
does all this make sense right it does we can find theta let’s go
so quadrant one or two but where is cosine positive this is a
positive number so we’re going to be in quadrant one
so here’s our angle theta right there and let’s see here we have cosine so
let’s think about this as one and then square root of five right here
and do we need any more sides well we’re looking for cosecant
so we’re going to need this missing side right here let’s call it h
so we have h squared plus 1 squared equals the square of square root of 5
which is 5. so h is 5 minus 1 which is 4 so h is in fact 2

00:28
positive 2 we’re going up all right so we have two right here now um
remember that cosecant of theta is one over sine of theta it’s just the
reciprocal of sine what is sine here sine is uh opposite over hypotenuse so
it’s going to be 2 over square root of 5
but this is the reciprocal so it’s going to be square root of 5 over 2. and
there we go so what we have here is cosecant of
cosine inverse 1 over square root of 5 all that is just equal to
square root of 5 over 2. so think of this right here as an angle
and we found cosecant of that angle square root of five over two
all right let’s do um now let’s do this without numbers can we
all right so here we go we’re going to take the same approach and see what we

00:29
can do so i’m going to think of this right here in inverse as an angle so
let’s call it theta and what do we have here right so we
need to switch the inputs and outputs so cosine of the angle is equal to x
and the angle is between 0 and pi now here’s where i’d like to see if this
makes sense or not this x here needs to be between -1 and 1 so you know [Music]
whatever answer we come up with we need to make sure that the
x here is always between minus one and one we don’t know what x is it’s just
something between minus one and one and theta is between 0 and pi now
before when we were drawing a diagram here we kind of didn’t
we kind of figured out what quadrant we’re in but because this x could be
positive or it could be negative right i mean we could put a 2 here or a minus 2

00:30
here so this could be a 2 or a -2 well that would be too large what i meant to
say is we could put a one-half here or a minus one-half year so we just don’t
know about this x so we don’t know if cosine is positive or cosine is negative
so i’ll just draw two two diagrams here so here’s one where cosine where theta
here is in quadrant one and here’s another triangle right here
where theta is in quadrant two in either case we have x over one
so we’re going to use x for the adjacent side here
whatever x is if x is positive or if x is negative
and then we’re going to use one for the hypotenuse so think of this as x over
one now we’re looking for sine of theta so we’re going to need opposite over
hypotenuse so we’re looking for h let’s go find h using pythagorean theorem so x
squared plus h squared equals 1 squared and so what will the h be

00:31
well h squared will be 1 minus x squared and then the h will be
plus or minus square root 1 minus x squared and so which one should we
change take plus or minus well no matter which diagram we’re using
the h is positive isn’t it right so we’re going to use h is 1 minus square
root of x squared right there this is what the h is going to be all right so
what we can say now is we’re looking for the sine of theta so sine of theta is
h over 1 which is just h so we got it so let’s just put it like this sine
n of cosine inverse of x is just the h and so that’s a nice thing to to see
sine of cosine inverse is just square root of 1 minus x squared

00:32
and that’s pretty sweet let’s put a box around that
all right let’s do uh one more here where it’s not quite as simple maybe a
little bit more let’s do sine of cosine inverse of 1 over x
what happens if we do this here all right so here we go
think of cosine inverse as an angle let’s call it theta
so theta is cosine inverse of 1 over x and what does that mean
it means that cosine theta is 1 over x and and the theta is between 0 and pi
all right so there we go let’s draw a triangle
because we’re looking for sine of theta and we don’t know sine of theta we know
cosine of theta so let’s draw a triangle now what quadrant can we be in this is

00:33
quadrant one or two and we don’t know what the x is so x
could be positive or x could be negative it can’t be zero but it could be
positive or negative so i’ll draw two triangles again one where theta is in
quadrant one because it can be less than or equal to pi and where theta is in
quadrant four sorry quadrant two in case theta is between zero and pi over two
or if it’s between pi over two and pi so we got two different scenarios there
but in in either scenario we’re going to say that the x is either a
one or a minus one and then this will be x
and then this will be minus x right up here so i drew one
situation for when x is greater than zero when it’s positive and i drew
another situation where x is less than zero now the diagram is really
just for us you don’t really actually need it to solve this problem we can say

00:34
what this is without drawing a diagram but i actually like to just draw a
diagram if this x is a positive like say a 3 then this will be the right diagram
if x is say negative 3 like if someone gave us 1 over negative 3
then this would be a negative 3 down here and we can say this is negative 1
and then this would be negative negative three remember the hypotenuse is always
positive so if your x is already negative you want to make sure and put a
negative four negative x all right in any case
uh what do we need we need sine of theta so we need the missing side right here
so here we go h squared plus one squared is x squared
so h squared is x squared minus one h is square root of x squared minus one
so now we can find out what sine is so sine of that’s what we’re asked for

00:35
right sine of theta so sine of cosine inverse of 1 over x is um
the opposite right here so we’re looking at sine sine so opposite over x so
square root of x squared over minus 1 over the hypotenuse x
so that’s what we’re looking for right there sine of cosine inverse
i guess we could put a box around that i was thinking about this problem and
you know because we just found sine um but i was thinking you know we can
actually find all six trig functions we don’t just need to find sine we could
find cosine and tangent and cotangent and i was thinking about those and what
happens if we do secant secant of this angle right here what is secant
and so secant is uh 1 over cosine right it’s just the reciprocal of cosine
cosine is what uh adjacent over hypotenuse so it’s 1

00:36
over x so this would be 1 over 1 over x which just reduces to x
so you know this leads to secant of theta so secant of cosine inverse 1 over x
and this is just x that’s really beautiful
secant of cosine inverse of 1 over x is x that’s a lot of fun all right um
in the next episode we’ll do some more inverse trig check it out right here

About The Author
Dave White Background Blue Shirt Squiggles Smile

David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

Let's do some math together, today!

Contact us today.

Account

Affiliates

About Us

Blog

© 2022 DAVE4MATH.com

Privacy Policy | Terms of Service