find the exact value of sin⁻¹(sin 7π/4). write your answer in radians in terms of π.

Explanation:

Step1: Recall the range of inverse - sine function

The range of (y = \sin^{-1}(x)) is (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]).

Step2: Rewrite (\sin\left(\frac{7\pi}{4}\right))

We know that (\sin\left(\frac{7\pi}{4}\right)=\sin\left(2\pi - \frac{\pi}{4}\right)). Using the identity (\sin(2\pi - \alpha)=-\sin\alpha), we have (\sin\left(\frac{7\pi}{4}\right)=-\sin\left(\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}).

Step3: Find (\sin^{-1}\left(\sin\left(\frac{7\pi}{4}\right)\right))

We want to find (\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)). Since the range of (y = \sin^{-1}(x)) is (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), and (\sin\left(-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}), then (\sin^{-1}\left(\sin\left(\frac{7\pi}{4}\right)\right)=-\frac{\pi}{4}).

Answer:

(-\frac{\pi}{4})