College Trigonometry, 2011a

822 Foundations of Trigonometry
(f) Since 11π
6
does not fall between 0 and π, Theorem10.26 does not apply. We are forced to
work through from the inside out starting with arccos ( cos ( )) ( √3 )
11π
6 = arccos
2
.From
( √3 )
the previous problem, we know arccos
2
= π 6 . Hence, arccos ( cos ( ))
11π
6 =
π
6 .
(g) One way to simplify cos ( arccos ( − 3 5))
is to use Theorem 10.26 directly. Since −
3
5 is
between −1 and 1, we have that cos ( arccos ( − 3 5))
= −
3
5
and we are done. However, as
before, to really understand why this cancellation occurs, we let t = arccos ( − 5) 3 . Then,
by definition, cos(t) =− 3 5 . Hence, cos ( arccos ( − 3 5))
= cos(t) =−
3
5
, and we are finished
in (nearly) the same amount of time.
(h) As in the previous example, we let t = arccos ( − 3 5)
so that cos(t) =−
3
5
for some t where
0 ≤ t ≤ π. Since cos(t) < 0, we can narrow this down a bit and conclude that π 2