Evaluate exactly.
-$$\cos ^ { - 1 } \left( \cos \left( - \frac { \pi } { 6 } \right) \right)$$
A) $\frac { 7 \pi } { 6 }$
B) $- \frac { 6 } { \pi }$
C) $\frac { \pi } { 6 }$
D) $- \frac { \pi } { 6 }$

Question

Quizplus · Accepted Answer

The cosine function is even, meaning $\cos(-x) = \cos(x)$. Therefore, $\cos^{-1}(\cos(-\frac{\pi}{6})) = \cos^{-1}(\cos(\frac{\pi}{6}))$, which simplifies directly to $\frac{\pi}{6}$ because $\frac{\pi}{6}$ is within the principal range of $\cos^{-1}$, which is $[0, \pi]$.

Evaluate Exactly cos⁡−1(cos⁡(−π6))\cos ^ { - 1 } \left( \cos \left( - \frac { \pi } { 6 } \right) \right)cos−1(cos(−6π​))

Evaluate Exactly $\cos ^ { - 1 } \left( \cos \left( - \frac { \pi } { 6 } \right) \right)$