Question

find the exact value of \\( \sin ^ { - 1 } \left( \sin \frac { 9 \pi } { 7 } \
ight) \\).
write your answer in radians in terms of \\( \pi \\).
if necessary, click on \undefined.\

Explanation:

Step1: Use the property of sine function

We know that \(\sin(x)=\sin(\pi - x)\). So, \(\sin\frac{9\pi}{7}=\sin(\pi+\frac{2\pi}{7})\). Since \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), here \(A=\pi\), \(B = \frac{2\pi}{7}\), \(\sin(\pi+\frac{2\pi}{7})=-\sin\frac{2\pi}{7}\).

Step2: Use the property of inverse - sine function

The range of \(y = \sin^{-1}(x)\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\). We want to find \(y=\sin^{-1}(\sin\theta)\). Let \(\theta=\frac{9\pi}{7}\). We rewrite \(\sin\frac{9\pi}{7}\) as \(-\sin\frac{2\pi}{7}=\sin(-\frac{2\pi}{7})\) (because \(\sin(-x)=-\sin x\)).
Since \(-\frac{\pi}{2}\leq-\frac{2\pi}{7}\leq\frac{\pi}{2}\), by the property \(\sin^{-1}(\sin x)=x\) for \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\)

Answer:

\(-\frac{2\pi}{7}\)