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Found in: Page 457

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 9–36, find the exact value of each expression.$\mathrm{tan}\left[{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right]$

The value of the expression $\mathrm{tan}\left[{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right]=-\frac{\sqrt{3}}{3}$.

See the step by step solution

## Step 1. Given information.

The given expression is:

$\mathrm{tan}\left[{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right]$

Let's take localid="1646413783497" $\theta ={\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)$

$\mathrm{sin}\theta =-\frac{1}{2}$

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

## Step 2. Find the value of θ.

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

By using the unit circle, we can say that $\theta$ must be equal to $-\frac{\mathrm{\pi }}{6}$, so that we can get $-\frac{1}{2}$.

$\mathrm{sin}\theta =-\frac{1}{2}\phantom{\rule{0ex}{0ex}}{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)=-\frac{\mathrm{\pi }}{6}$

## Step 3. Find the exact value of the expression.

$\mathrm{tan}\left(-\frac{\mathrm{\pi }}{6}\right)=-\frac{\sqrt{3}}{3}$

Therefore, role="math" localid="1646414553456" $\mathrm{tan}\left[{\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)\right]=-\frac{\sqrt{3}}{3}$.