QuestionFind the value of $\sin ^{-1}\left[\sin \left(\frac{3 \pi}{5}\right)\right]$ .

Studdy Solution

STEP 1

Assumptions1. We are working in radians, not degrees. . The inverse sine function, $\sin^{-1}(x)$ , has a range of $[-\frac{\pi}{}, \frac{\pi}{}]$ .
3. The sine function, $\sin(x)$ , has a period of $\pi$ .

STEP 2

First, note that the expression inside the inverse sine function is a sine function. So, we are taking the inverse sine of a sine function.
$\sin^{-1}\left[\sin\left(\frac{\pi}{5}\right)\right]$

STEP 3

The sine function, $\sin(x)$ , and the inverse sine function, $\sin^{-1}(x)$ , are inverse functions of each other. Therefore, for any $x$ in the domain of $\sin(x)$ , we have $\sin^{-1}(\sin(x)) = x$ .
However, this is true only if $x$ is in the range of $\sin^{-1}(x)$ , which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ .

STEP 4

We check if $\frac{3\pi}{}$ is in the range of $\sin^{-1}(x)$ .
Since $\frac{3\pi}{}$ is greater than $-\frac{\pi}{2}$ and less than $\frac{\pi}{2}$ , we can say that $\frac{3\pi}{}$ is in the range of $\sin^{-1}(x)$ .

STEP 5

Therefore, we can simplify the given expression as follows $\sin^{-1}\left[\sin\left(\frac{3\pi}{5}\right)\right] = \frac{3\pi}{5}$ So, the solution to the problem is $\frac{3\pi}{5}$ .

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