Question

\( \cos^{-1}\left(\cos\left(\frac{7\pi}{5}\right)\right) = \)

• A. \( \frac{3\pi}{5} \)
• B. \( \frac{2\pi}{5} \)
• C. \( \frac{-7\pi}{5} \)
• D. \( \frac{7\pi}{5} \)
• E. \( \frac{-2\pi}{5} \)

Hint:

For \( \cos^{-1}(\cos \theta) \), if \( \theta \) is in the 3rd or 4th quadrant, use the formula \( 2\pi - \theta \) to bring it back to the principal range.

Answer 1

The Correct Option is A

Concept: The range of the principal value branch of \( \cos^{-1}x \) is \( [0, \pi] \). If the angle \( \theta \) is outside this range, we must use the periodicity and symmetry properties of the cosine function (\( \cos\theta = \cos(2\pi - \theta) \)) to find an equivalent angle within \( [0, \pi] \).

Step 1: Check the range.
The given angle is \( \theta = \frac{7\pi}{5} \). Since \( \frac{7\pi}{5} > \pi \), it is outside the principal range of \( \cos^{-1} \).

Step 2: Use symmetry properties.
We know \( \cos(2\pi - \alpha) = \cos \alpha \). Let's find \( 2\pi - \frac{7\pi}{5} \): \[ 2\pi - \frac{7\pi}{5} = \frac{10\pi - 7\pi}{5} = \frac{3\pi}{5} \]

Step 3: Verify the new angle.
Is \( \frac{3\pi}{5} \in [0, \pi] \)? Yes, because \( 0 < 0.6\pi < \pi \). Therefore: \[ \cos^{-1}\left(\cos\frac{7\pi}{5}\right) = \cos^{-1}\left(\cos\frac{3\pi}{5}\right) = \frac{3\pi}{5} \]