The value of sin^-1(sin 10) , where 10 is in radians, is

The correct answer is Option (3) → \(3\pi - 10\)

The answer must be the angle in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) whose sine equals \(\sin 10\). Taking \(\pi \approx 3.1416\), first we locate 10 between multiples of \(\pi\):

    \(3\pi \approx 9.42\) and \( \displaystyle \frac{7\pi}{2} \approx 11.00 \)

So \(10 = 3\pi + t\) with \(t = 10 - 3\pi \approx 0.58\), a small positive angle. Since \(\sin(3\pi + t) = -\sin t\),

    \(\sin 10 = -\sin(10 - 3\pi)\)

    \(= \sin(3\pi - 10)\)

Now \(3\pi - 10 \approx -0.58\), which lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Hence

    \(\sin^{-1}(\sin 10) = 3\pi - 10\)

Why the other options are incorrect

• Option (1) \(10 - 3\pi\) — Incorrect. \(10 - 3\pi\) is positive, so its sine is positive, but \(\sin 10\) is negative (about \(-0.54\)).

• Option (2) \(4\pi - 10\) — Incorrect. \(4\pi - 10 \approx 2.57\), which is greater than \(\frac{\pi}{2}\).

• Option (4) \(10\) — Incorrect. 10 is far outside \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).