Question

Find \( \cos^{-1} \left( \cos \frac{8\pi}{5} \right) \).

• A. \( \frac{8\pi}{5} \)
• B. \( \frac{2\pi}{5} \)
• C. \( \frac{\pi}{5} \)
• D. \( \frac{3\pi}{5} \)

Hint:

For angles greater than \( \pi \), subtract multiples of \( 2\pi \) to bring the angle within the principal range of \( \cos^{-1} \).

Answer 1

The Correct Option is B

We are asked to find: \[ \cos^{-1} \left( \cos \frac{8\pi}{5} \right) \] The principal range of \( \cos^{-1} x \) is \( [0, \pi] \). Since \( \frac{8\pi}{5} \) is greater than \( \pi \), we need to find an equivalent angle in the range \( [0, \pi] \). The angle \( \frac{8\pi}{5} \) is greater than \( \pi \), so we subtract \( 2\pi \) from it: \[ \frac{8\pi}{5} - 2\pi = \frac{8\pi}{5} - \frac{10\pi}{5} = -\frac{2\pi}{5} \] Now, since \( \cos \) is an even function, we know that: \[ \cos \left( -\frac{2\pi}{5} \right) = \cos \left( \frac{2\pi}{5} \right) \] Therefore, we have: \[ \cos^{-1} \left( \cos \frac{8\pi}{5} \right) = \frac{2\pi}{5} \] Thus, the correct answer is \( \frac{2\pi}{5} \).