Evaluate the expressions.
Find the exact values.
Here we have the inverse cosine of cosine
of negative 37 10ths pi radians.
Let's begin by sketching this angle
in the standard position.
Notice that negative 37 10ths pi radians
is equal to negative three and seven 10ths pi radians.
If the angle is negative four pi radians,
we would have two complete rotations
in the clockwise direction,
and because our angle is negative three
and seven 10ths pi radians,
notice our angle is going to be three 10ths pi radians
short of two complete rotations in the clockwise direction.
So in sketching the angle, we begin
along the positive X axis.
Now we rotate clockwise
three and seven 10ths pi radians.
We have negative one pi, negative two pi,
negative three pi.
We need to rotate another seven 10ths pi radians
in the clockwise direction,
and since seven 10ths pi radians
is more than one half pi radians, or pi over two radians,
this will bring us back into the first quadrant.
We rotate another seven 10ths pi radians
in the clockwise direction,
which will bring us to approximately here,
where again, this is three 10ths pi radians
short of two complete rotations in the clockwise direction,
which is important because this tells us
the reference angle that's acute angle here
is three 10ths pi radians.
And so when this is a unit circle,
the terminal side intersects the unit circle at this point.
Let's call the ordered pair for this point X sub one comma
Y sub one, where on the unit circle,
X equals cosine theta and Y equals sine theta,
which means we can represent cosine
of negative 37 10ths pi radians as X sub one.
Let's go ahead and do that.
This is equal to inverse cosine of X sub one.
Inverse cosine of X sub one
will return an angle
in the interval from 0 to pi radians,
including the endpoints,
that has a cosine function value of X sub one,
so the angle must be in the interval from 0 to pi radians,
including the endpoints.
We know any angle that has a terminal side of this ray
will have a cosine function value of X sub one,
and notice the reference angle of three 10ths pi radians,
this angle here,
will have a cosine function value of X sub one
and the angle is in the interval from 0 to pi radians,
and therefore, this expression simplifies
to three 10ths pi radians.
Let's take a look at one more example.
We have the inverse sine of sine
of negative 36 11ths pi radians.
Again, let's sketch the angle in standard position.
Notice that negative 36 11ths pi
is equal to negative three
and three 11ths pi.
In standard position, we rotate clockwise
three and three 11ths pi radians,
so we have negative one pi,
negative two pi, negative three pi.
We need to rotate another three 11ths pi radians
past the negative X axis,
which brings us to approximately here.
This would be the terminal side
of negative 36 11ths pi radians.
Because we rotated an additional three 11ths pi radians
past the negative X axis,
we now know the reference angle of this angle here
is three 11ths pi radians.
We are in the second quadrant,
so let's let the ordered pair for this point
here on the unit circle be negative X sub two
comma Y sub two, because in the second quadrant,
X is negative and Y is positive.
And since sine theta is equal to Y,
we can represent the sine of negative 36 11ths pi radians
as Y sub two.
Let's write the expression as inverse sine
of Y sub two.
Inverse sine of Y sub two
will return an angle
in the interval from negative pi over two
to positive pi over two, including the endpoints,
that has a sine function value of Y sub two,
which means the angle must be in the interval
from 0 to pi over two
or from 0 to negative pi over two.
Notice how the terminal side of the angle
is not in the correct quadrant, but remember,
sine is positive in the first quadrant
as well as in the second quadrant,
so if we sketch the same reference angle
in the first quadrant,
we can find the angle that we need to evaluate
inverse sine of Y sub two.
So the terminal side in quadrant one
that will give us a reference angle
of three 11ths pi radians, would be this ray here,
where again, the reference angle is still
three 11ths pi radians.
This point on the unit circle in the first quadrant
would be X sub two comma Y sub two
because both X and Y are positive.
But notice how, because the Y coordinate is still Y sub two,
the sine function value is still Y sub two
for any angle that has this ray as its terminal side,
and therefore the angle in the interval
from negative pi over two to positive pi over two
that has a sine function value of Y sub two
is this reference angle of three 11ths pi radians.
The given expression simplifies to three 11ths pi radians.
I hope you found this helpful.