Question

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

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

Short answer

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

Step-by-step solution

Step 1. Given information. 

The given expression is:

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

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

$\sin \theta =\left(-\frac{1}{2}\right)$

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

Step 2. Find the value of θ.

Initially, we have function ${\sin }^{-1}$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 $-\frac{\mathrm{\pi }}{6}$, 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.

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

By using the unit circle, we know that $\cos \left(-\frac{\mathrm{\pi }}{6}\right)=\frac{\sqrt{3}}{2}$, then

localid="1646454996847" $$\begin{gathered} \sec \left(-\frac{\mathrm{\pi }}{6}\right)=\frac{1}{{\displaystyle \frac{\sqrt{3}}{2}}} \\ \sec \left(-\frac{\mathrm{\pi }}{6}\right)=\frac{2}{\sqrt{3}} \\ \sec \left(-\frac{\mathrm{\pi }}{6}\right)=\frac{2}{\sqrt{3}}·\frac{\sqrt{3}}{\sqrt{3}} \\ \sec \left(-\frac{\mathrm{\pi }}{6}\right)=\frac{2\sqrt{3}}{3} \end{gathered}$$

Therefore,$\sec \left[{\sin }^{-1}\left(-\frac{1}{2}\right)\right]=\frac{2\sqrt{3}}{3}$.