College Trigonometry, 2011a

10.6 The Inverse Trigonometric Functions 821
Everything in Theorem 10.26 is a direct consequence of the facts that f(x) = cos(x) for 0 ≤ x ≤ π
and F (x) = arccos(x) are inverses of each other as are g(x) = sin(x) for− π 2 ≤ x ≤ π 2 and
G(x) = arcsin(x). It’s about time for an example.
Example 10.6.1.
1. Find the exact values of the following.
(a) arccos ( )
1
2
(
(c) arccos −
√ )
2
2
(e) arccos ( cos ( ))
π
6
(g) cos ( arccos ( − 3 ))
5
(b) arcsin
( √2
2
(d) arcsin ( − 1 2
)
(f) arccos ( cos ( ))
11π
6
(h) sin ( arccos ( − 3 ))
5
)
2. Rewrite the following as algebraic expressions of x and state the domain on which the equivalence
is valid.
(a) tan (arccos (x))
(b) cos (2 arcsin(x))
Solution.
1. (a) To find arccos ( 1
2)
, we need to find the real number t (or, equivalently, an angle measuring
t radians) which lies between 0 and π with cos(t) = 1 2 . We know t = π 3
meets these
criteria, so arccos ( )
1
2 =
π
3
( .
√2 )
(b) The value of arcsin
2
is a real number t between − π 2 and π √
2 with sin(t) = 2
2
( . The
√2 )
number we seek is t = π 4 . Hence, arcsin 2
= π 4 .
( √ )
√
− 2
2
lies in the interval [0,π] with cos(t) =− 2
2
. Our answer
(c) The number t = arccos
( )
is arccos − = 3π 4 .
√
2
2
(d) To find arcsin ( − 1 [
2)
, we seek the number t in the interval −
π
2 , π ]
2 with sin(t) =−
1
2 .The
answer is t = − π 6 so that arcsin ( − 1 2)
= −
π
6 .
(e) Since 0 ≤ π 6 ≤ π, we could simply invoke Theorem 10.26 to get arccos ( cos ( ))
π
6 =
π
6 .
However, in order to make sure we understand why this is the case, we choose to work
the example through using the definition of arccosine. Working from the inside out,
arccos ( cos ( )) ( √3 )
( √3 )
π = arccos . Now, arccos is the real number t with 0 ≤ t ≤ π
and cos(t) =
6
2
√
3
2 . We find t = π 6 , so that arccos ( cos ( π
6
2
))
=
π
6 .