## Precalculus (6th Edition) Blitzer

The exact value of the provided expression is $\frac{\pi }{2}$.
Let $\theta ={{\sin }^{-1}}1$. Then, we have $\sin \theta =1$ For the sine function, the interval for the angle is $\left[ -\frac{\pi }{2},\frac{\pi }{2} \right]$. So, the only angle that satisfies $\sin \theta =1$ is $\frac{\pi }{2}$. Thus, $\theta =\frac{\pi }{2}$ and the exact value of given expression is ${{\sin }^{-1}}1=\frac{\pi }{2}$