Question

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

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

Short answer

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

Step-by-step solution

Step 1. Given information.

The given expression is:

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

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

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

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

Step 2. Find the value of θ.

Initially, we have 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 must $\theta$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.

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

As we know $\tan \theta =\frac{\sin \theta }{\cos \theta }$, then

localid="1646422050104" $$\begin{gathered} \tan \left(-\frac{\mathrm{\pi }}{6}\right)=\frac{\sin \left(-\frac{\mathrm{\pi }}{6}\right)}{\cos \left(-\frac{\mathrm{\pi }}{6}\right)} \\ =-\frac{\sin \left(\frac{\mathrm{\pi }}{6}\right)}{\cos \left(\frac{\mathrm{\pi }}{6}\right)} \\ =-\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}} \\ =-\frac{1}{\sqrt{3}} \\ \tan \left(-\frac{\mathrm{\pi }}{6}\right)=-\frac{1}{\sqrt{3}} \\ \cot \left(-\frac{\mathrm{\pi }}{6}\right)=\frac{1}{\left(-\frac{1}{\sqrt{3}}\right)} \\ =-\sqrt{3} \end{gathered}$$

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