Calculus BC
Inverse Trigonometry
Part 3C Notes
Example:
cos1 (cos  )  
cos(cos 1  )  does not exist
cos 1 (cos 53 )  3
It is important to recognize that the range of the sine and cosine functions is [–1, 1]
and the domain of the inverse sine and inverse cosine functions is likewise [–1, 1].
The range of inverse sine is [  2 , 2 ] and the range of the inverse cosine function is
[ 0 ,  ]. The inverse tangent function has [  2 , 2 ] as its range.
Example:
Write as an algebraic function of x: tan(sin 1 2 x)
The inside argument, sin 1 (2 x) is some angle,  . Namely,   sin 1 (2 x) , or after
taking the sine of both sides, sin   2x for some angle  . Use that to sketch a
triangle.
1
2x

We got the above triangle from the fact that sine is opposite over hypotenuse.
Now find the missing adjacent side of the triangle above using a 2  b 2  c 2 .
The triangle complete, is:
1
2x

1  4x 2
So we can now realize that we wanted tan(sin 1 2 x) but that is just tan 
(looking at the definition of  above for this problem) so we can make use of the
fact that tangent is opposite over adjacent to get
tan(sin 1 2 x) 
Example:
2x
1  4 x2
. There is no need to waste time rationalizing.
Notice that this same technique (above) also works for getting numeric values. If
you had been asked for sec(tan 1 23 ) you would have set up a triangle where the
tangent of the angle was
2
3
, as shown at the top of the next page.
2

3
Then found the missing side (here it is the hypotenuse that is missing) using the
Pythagorean Theorem.
In this example the hypotenuse is 11 and use the definition of cosine (adjacent
over hypotenuse) to get the final answer:
sec(tan 1 23 ) 
Example:
11
.
3
Find the value of cos(cos 1 135  tan 1 12 )
This one first requires that we realize that both cos 1 135 and tan 1 12 are angles.
Whenever we use an inverse trig function, we are getting angles. So we can treat
the problem like we are taking cos(   ) where  and  are angles. In other
words, use the sum formula for cosine.
cos( A  B)  cos A cos B  sin Asin B .
Therefore,
cos(cos 1 135  tan 1 12 )  cos(cos1 135 )  cos(tan 1 12 )  sin(cos1 135 )  sin(tan 1 12 )
Wow.
This is basically like four problems in one. Aside from the very first term where
cosine and inverse cosine cancel each other out (since cos(cos 1 135 )  135 ) you
basically have to do work just like the previous example three times over,
working out:
cos(tan 1 12 ) = you go figure it out
sin(cos 1 135 ) = go figure this one out too, and
sin(tan 1 12 ) = more work for students to do, ha ha ha!
Once you do that, you get a final answer of:
cos(cos 1 125  tan 1 12 )
=
5 2 12 1

 
13 5 13 5
= 
2
13 5