Question

\[ \cos^{-1} \left( \cos \left( \frac{-7\pi}{9} \right) \right) = \]

• A. \(\frac{-7\pi}{9}\)
• B. \(\frac{7\pi}{9}\)
• C. \(\frac{2\pi}{9}\)
• D. \(\frac{-2\pi}{9}\)
• E. \(\frac{-4\pi}{9}\)

Hint:

Always remember that \(\cos^{-1}(x)\) refers to the principal value, which for cosine lies between 0 and \(\pi\). This is crucial when angles fall outside this range and need to be adjusted accordingly.

Answer 1

The Correct Option is B

The function \(\cos^{-1}(x)\) returns the principal value, which must be in the range \([0, \pi]\). For any \(\theta\), \(\cos(\theta)\) is periodic with a period of \(2\pi\) and symmetric about the y-axis. 
Therefore, \(\cos(\theta) = \cos(-\theta)\).
Considering \(\theta = \frac{-7\pi}{9}\), the equivalent angle in the principal range can be found by converting \(\theta\) to a positive angle, as the cosine function is even: \[ \cos\left(\frac{-7\pi}{9}\right) = \cos\left(\frac{7\pi}{9}\right) \] The angle \(\frac{7\pi}{9}\) lies within the principal range \([0, \pi]\). Thus, the principal value of \(\cos^{-1}\) applied to \(\cos\left(\frac{-7\pi}{9}\right)\) is: \[ \cos^{-1}\left(\cos\left(\frac{7\pi}{9}\right)\right) = \frac{7\pi}{9} \]