Question

The principal value of \( \sin^{-1} \left( \cos \frac{43\pi}{5} \right) \) is

• A. \( -\frac{7\pi}{5} \)
• B. \( -\frac{\pi}{10} \)
• C. \( \frac{\pi}{10} \)
• D. \( \frac{3\pi}{5} \)

Hint:

Remember the principal value range of \( \sin^{-1} x \) is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). To solve such problems, first simplify the angle inside the trigonometric function using periodicity. Then, use trigonometric identities to convert the outer function's argument into a sine function. Finally, ensure the resulting angle is within the principal value range.

Answer 1

The Correct Option is B

We need to find the principal value of \( \sin^{-1} \left( \cos \frac{43\pi}{5} \right) \). First, we simplify the angle \( \frac{43\pi}{5} \): $$ \frac{43\pi}{5} = 8\pi + \frac{3\pi}{5} $$ Since the cosine function has a period of \( 2\pi \), we have: $$ \cos \left( \frac{43\pi}{5} \right) = \cos \left( \frac{3\pi}{5} \right) $$ Now, we use the identity \( \cos x = \sin \left( \frac{\pi}{2} - x \right) \): $$ \cos \left( \frac{3\pi}{5} \right) = \sin \left( \frac{\pi}{2} - \frac{3\pi}{5} \right) $$ $$ \frac{\pi}{2} - \frac{3\pi}{5} = \frac{5\pi - 6\pi}{10} = -\frac{\pi}{10} $$ So, we have: $$ \cos \left( \frac{3\pi}{5} \right) = \sin \left( -\frac{\pi}{10} \right) $$ Now we need to find the principal value of \( \sin^{-1} \left( \sin \left( -\frac{\pi}{10} \right) \right) \). The principal value range for \( \sin^{-1} x \) is \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). Since \( -\frac{\pi}{10} \) lies within this range, the principal value is \( -\frac{\pi}{10} \).