# 005 Sample Final A, Question 13

Question Give the exact value of the following if its defined, otherwise, write undefined.
${\displaystyle (a)\sin ^{-1}(2)\qquad \qquad (b)\sin \left({\frac {-32\pi }{3}}\right)\qquad \qquad (c)\sec \left({\frac {-17\pi }{6}}\right)}$

Foundations:
1) What is the domain of ${\displaystyle \sin ^{-1}?}$
2) What are the reference angles for ${\displaystyle {\frac {-32\pi }{3}}}$ and ${\displaystyle {\frac {-17\pi }{6}}}$?
1) The domain is ${\displaystyle [-1,1].}$
2) The reference angle for ${\displaystyle {\frac {-32\pi }{3}}}$ is ${\displaystyle {\frac {4\pi }{3}}}$, and the reference angle for ${\displaystyle {\frac {-17\pi }{6}}}$ is ${\displaystyle {\frac {7\pi }{6}}}$

Step 1:
For (a), we want an angle ${\displaystyle \theta }$ such that ${\displaystyle \sin(\theta )=2}$. Since ${\displaystyle -1\leq \sin(\theta )\leq 1}$, it is impossible
for ${\displaystyle \sin(\theta )=2}$. So, ${\displaystyle \sin ^{-1}(2)}$ is undefined.
Step 2:
For (b), we need to find the reference angle for ${\displaystyle {\frac {-32\pi }{3}}}$. If we add multiples of ${\displaystyle 2\pi }$ to this angle, we get the
reference angle ${\displaystyle {\frac {4\pi }{3}}}$. So, ${\displaystyle \sin \left({\frac {-32\pi }{3}}\right)=\sin \left({\frac {4\pi }{3}}\right)={\frac {-{\sqrt {3}}}{2}}}$.
Step 3:
For (c), we need to find the reference angle for ${\displaystyle {\frac {-17\pi }{6}}}$. If we add multiples of ${\displaystyle 2\pi }$ to this angle, we get the
reference angle ${\displaystyle {\frac {7\pi }{6}}}$. Since ${\displaystyle \cos \left({\frac {7\pi }{6}}\right)={\frac {-{\sqrt {3}}}{2}}}$, we have
${\displaystyle \sec \left({\frac {-17\pi }{6}}\right)=\sec \left({\frac {7\pi }{6}}\right)={\frac {2}{-{\sqrt {3}}}}={\frac {-2{\sqrt {3}}}{3}}}$.
b) ${\displaystyle {\frac {-{\sqrt {3}}}{2}}}$
c)${\displaystyle {\frac {-2{\sqrt {3}}}{3}}}$