Inverse trig functions: composition

OK.

Let's take a look at how to solve

some composition of trig functions with their inverses.

So, all of these functions today are going to be

a trig function inverse of a trig function of an angle.

Often when you evaluate composition of functions in the inverses,

it's very easy.

You just take whatever that input is as your answer.

That's not going to work here because we've restricted

the range of

the inverse trig functions so much.

So,

these first three examples we have the same angle 5 pi over 8.

Let's see where that is on a unit circle.

It's a good idea whenever you're doing these to

draw a circle

and

draw in the angle.

That's about approximately 5 pi over 8.

We can't say that sine inverse of sine of 5 pi over 8 is just 5 pi over 8 because

our answer to any sine inverse question needs to be in the 4th or 1st quadrant,

specifically between negative pi over 2 and positive pi over 2.

So, we need to figure out which angle

has the same sine value as 5 pi over 8.

So, if we just come straight across,

reflect over the y-axis,

that will give us a point that has the same y

value. Place on the unit circle with the same y value.

We just want to figure out

what is this angle here.

Well,

to get to 5 pi over 8,

we could have started on the negative

x-axis and rotated backwards

3/8 of pi because 5 is 3 from 8.

So, let's just go forward 3 pi over 8

from the positive x-axis so we get 3 pi

over 8.

Tangent inverse of tangent 5 pi over 8

is not going to be the same thing.

We need to find a place in the unit circle where y over x is the same

as it is at

this point up here where the angle is 5 pi over 8.

We need therefore to be in the 4th quadrant because

tangent is negative in the 2nd quadrant and also in the 4th quadrant.

So, what we're going to do is we're going to actually rotate

pi radians a 180 degrees around.

So,

this angle here must be negative

3 pi over 8.

I figured that out by again saying

I'm 3 pi over 8 away from the negative x-axis,

so I need a backup 3 pi over 8 from the positive x-axis.

What about cosine inverse of cosine of 5 pi over 8?

What do we need to change there?

Well,

turns out nothing.

Unlike sine and tangent inverse,

cosine inverse

is

always going to give us an answer in the first or second quadrant.

5 pi over 8 is already in the second quadrant,

so we don't need to change anything at all.

OK.

What about

this second angle 12 pi over 7?

Let's draw that in.

That's going to be in the 4th quadrant,

maybe about right there.

And

inverse of sine of 12 pi over 7,

you might think it's just 12 pi over 7 because we're already in the fourth quadrant.

But we need to name the angle in such a way that the

angle is between negative pi over 2 and positive pi over 2.

So, we don't have to change the position on the unit circle.

We're in the right place.

We need to give it the right name, and the right name in this case

is negative 2 pi over 7.

I know that because

12 pi over 7 is just 2/7

away from being

14 pi over 7,

which would be 2 pi a complete revolution.

How about tangent inverse of tangent of 12 pi over 7?

Well,

same deal.

We're in the correct place.

We don't need to rotate at all or reflect over the y-axis.

So, we just need to give it the right name,

which is -2 pi over 7.

But what about cosine inverse of cosine 12 pi over 7?

Here we do have to do something.

We have to change the point

because we're not in the first or second quadrant.

How do we get to the right place?

Well,

we want to find a place in the unit circle that is the same x value.

So, let's go straight on up,

and it must be right around there.

So, what is this angle here?

That angle there,

we went backwards 2 pi over 7 to get down to 12 pi over 7.

So, let's go forwards

the same amount,

so this will be positive 2 pi over 7.

OK.

So that's how you solve those.

Why don't you try a couple on your own here?

Let's move this over.

I have a few for you to try out.

Try those.

Pause the video,

try those,

and then I'll come back and tell you if you have the right answer.

All right,

let's take a look.

6 pi over 5.

Good idea to quickly sketch a circle,

unit circle and see where is that

6 pi over 5.

It's just over 5 pi over 5,

which is half a circle.

So, we're right around there.

Now,

we're in the 3rd quadrant,

so we do need to change the position of the point

for

sine inverse.

We want to reflect over the y-axis,

so we'll have the same y value.

So, we need to be right there.

So, instead of going 1/5 of pi beyond the x-axis,

we're going to come back a 5th of pi.

This should be negative

pi over 5.

For tangent,

we also need to change,

but now we want to be in the first quadrant because the

tangent will be positive like it is in the 3rd quadrant.

So, in this case, it will be positive pi over 5.

And for cosine inverse of cosine of 6 pi over 5,

again we need to change the position because

we're not in the first or second quadrant.

So, we'll reflect over the x-axis that'll keep.

It'll keep the x value the same.

And so, now instead of going 1 more fifth of pi from the negative x-axis,

we need a backup one.

So, instead of 5 pi we go back one,

that would be

4

pi over 5.

All right,

that's how you solve problems like this.

I hope this has helped,

and thanks for watching.