Question

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

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

Short answer

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

Step-by-step solution

Step 1. Given information.

The given expression is:

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

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

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

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

Step 2. Find the value of θ.

Initially, we have${\sin }^{-1}$ and the bounds of sinfunction 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 $-\frac{\mathrm{\pi }}{6}$, so that we can get $-\frac{1}{2}$.

$$\begin{gathered} \sin \theta =-\frac{1}{2} \\ {\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\mathrm{\pi }}{6} \end{gathered}$$

Step 3. Find the exact value of the expression.

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

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