Question

find the exact value of \\( \sin ^ { - 1 } \left( \sin \frac { 5 \pi } { 6 } \
ight) \\).
write your answer in radians in terms of \\( \pi \\).

Explanation:

Step1: Recall the property of inverse sine function

The property of \(y = \sin^{-1}(x)\) is that \(\sin^{-1}(\sin\theta)=\theta\) when \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\).
First, find \(\sin\frac{5\pi}{6}\). We know that \(\sin\frac{5\pi}{6}=\sin(\pi - \frac{\pi}{6})\). Using the formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\) (here \(A=\pi\), \(B = \frac{\pi}{6}\)), \(\sin\pi\cos\frac{\pi}{6}-\cos\pi\sin\frac{\pi}{6}=0\times\frac{\sqrt{3}}{2}-(- 1)\times\frac{1}{2}=\frac{1}{2}\).

Step2: Find the value of \(\sin^{-1}(\sin\frac{5\pi}{6})\)

Let \(\theta=\frac{5\pi}{6}\), but \(\frac{5\pi}{6}
otin[-\frac{\pi}{2},\frac{\pi}{2}]\). We know that \(\sin\frac{5\pi}{6}=\sin(\pi-\frac{\pi}{6})=\sin\frac{\pi}{6}\), and \(\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}]\).
Since \(\sin^{-1}(\sin x)=x\) for \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\), when \(x = \frac{\pi}{6}\), \(\sin^{-1}(\sin\frac{5\pi}{6})=\sin^{-1}(\sin\frac{\pi}{6})\)

Answer:

\(\frac{\pi}{6}\)