College Trigonometry, 2011a

10.6 The Inverse Trigonometric Functions 829
Solution.
1. (a) Using Theorem 10.28, we have arcsec(2) = arccos ( )
1
2 =
π
3 .
(b) Once again, Theorem 10.28 comes to our aid giving arccsc(−2) = arcsin ( − 1 2)
= −
π
6 .
(c) Since 5π 4
doesn’t fall between 0 and π 2 or π 2
and π, we cannot use the inverse property
stated in Theorem 10.28. We can, nevertheless, begin by working ‘inside out’ which
yields arcsec ( sec ( )) √ ( √ )
5π
4 = arcsec(− 2) = arccos − 2
2
= 3π 4 .
(d) One way to begin to simplify cot (arccsc (−3)) is to let t = arccsc(−3). Then, csc(t) =−3
and, since this is negative, we have that t lies in the interval [ − π 2 , 0) . We are after
cot (arccsc (−3)) = cot(t), so we use the Pythagorean Identity 1 + cot 2 (t) = csc 2 (t).
Substituting, we have 1 + cot 2 (t) =(−3) 2 , or cot(t) =± √ 8=±2 √ 2. Since − π 2 ≤ t