find the following without using a calculator.\n$cos(sin^{-1}\frac{sqrt{2}}{2})=\frac{sqrt{?}}{}$

Explanation:

Step1: Let $\theta=\sin^{-1}\frac{\sqrt{2}}{2}$

By the definition of the inverse - sine function, $\sin\theta=\frac{\sqrt{2}}{2}$ and $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$.

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

We know $\sin\theta=\frac{\sqrt{2}}{2}$, so $\left(\frac{\sqrt{2}}{2}\right)^{2}+\cos^{2}\theta = 1$. Then $\frac{2}{4}+\cos^{2}\theta=1$, which simplifies to $\cos^{2}\theta=1 - \frac{1}{2}=\frac{1}{2}$.

Step3: Determine the sign of $\cos\theta$

Since $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$ and $\sin\theta=\frac{\sqrt{2}}{2}>0$, $\theta$ is in the first - quadrant where $\cos\theta>0$. So $\cos\theta=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$.

Answer:

$\frac{\sqrt{2}}{2}$