The value of sin^-1≤ft(cos(9π)/7) is

The correct answer is Option (2) → \(-\displaystyle\frac{3\pi}{14}\)

The cosine has to be written as the sine of an angle in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), the principal value branch of \(\sin^{-1}\).

Since \(\frac{9\pi}{7} = \pi + \frac{2\pi}{7}\) is in the third quadrant, its cosine is negative:

    \( \displaystyle \cos\frac{9\pi}{7} = -\cos\frac{2\pi}{7} \)

Using \(\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)\):

    \( \displaystyle \cos\frac{2\pi}{7} = \sin\left(\frac{\pi}{2} - \frac{2\pi}{7}\right) = \sin\frac{3\pi}{14} \)

So

    \( \displaystyle \cos\frac{9\pi}{7} = -\sin\frac{3\pi}{14} = \sin\left(-\frac{3\pi}{14}\right) \)

Since \(-\frac{3\pi}{14}\) lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\),

    \( \displaystyle \sin^{-1}\left(\cos\frac{9\pi}{7}\right) = -\frac{3\pi}{14} \)

Why the other options are incorrect

• Option (1) \(\frac{3\pi}{14}\) — Incorrect. \(\sin\frac{3\pi}{14}\) is positive, but \(\cos\frac{9\pi}{7}\) is negative.

• Option (3) \(-\frac{2\pi}{7}\) — Incorrect. \(\frac{2\pi}{7}\) is the angle inside the cosine; the sine of \(-\frac{2\pi}{7}\) is about \(-0.78\), while \(\cos\frac{9\pi}{7}\) is about \(-0.62\).

• Option (4) \(\frac{9\pi}{14}\) — Incorrect. \(\frac{9\pi}{14}\) is greater than \(\frac{\pi}{2}\), so it is outside the principal branch.