Question

In Problems 9–36, find the exact value of each expression.

$\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)$

Short answer

The value of the expression$\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}$.

Step-by-step solution

Step 1. Given information.

The given expression is:

$\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)$

Let's take $\theta ={\sin }^{-1}\frac{\sqrt{2}}{2}$,

$-\frac{\mathrm{\pi }}{2}\le \theta \le \frac{\mathrm{\pi }}{2}$is the domain of the function.

Step 2. Find the value of θ.

Initially, we have ${\sin }^{-1}$and the bounds of sine function are going to be in the interval$\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$. It represents the range of ${\sin }^{-1}$.

By using the unit circle, we can say that $\theta$must be equal to $\frac{\mathrm{\pi }}{4}$, so that we can get $\frac{\sqrt{2}}{2}$.

$$\begin{gathered} \sin \theta =\frac{\sqrt{2}}{2} \\ {\sin }^{-1}\frac{\sqrt{2}}{2}=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Step 3. Find the exact value of the expression.

$\cos \left(\frac{\mathrm{\pi }}{4}\right)=\frac{\sqrt{2}}{2}$

Therefore,$\cos \left({\sin }^{-1}\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}$.