Chapter 7, Problem 26RE

To determine

To Calculate: The simplified value of the expression $\tan \left({\cos }^{-1}\left(-\frac{1}{2}\right)\right)$ .

Answer to Problem 26RE

The simplified value of the expression $\tan \left({\cos }^{-1}\left(-\frac{1}{2}\right)\right)$ is $\sqrt{3}$ .

Explanation of Solution

Given information:

The expression is $\tan \left({\cos }^{-1}\left(-\frac{1}{2}\right)\right)$ .

Formula used:

For the cosine function and its inverse, the following holds:

  $\begin{array}{c}f\left(f^{-1}\left(x\right)\right)=\cos \left({\cos }^{-1}\left(x\right)\right)\\ =x\end{array}$

Where, $-1\le x\le 1$

Tangent function and cosine function related as,

  $\tan \theta =\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }$

Calculation:

Consider the expression $\tan \left({\cos }^{-1}\left(-\frac{1}{2}\right)\right)$ .

Simplify it,

  $\tan \left({\cos }^{-1}\left(-\frac{1}{2}\right)\right)=\tan \left(\pi -{\cos }^{-1}\left(\frac{1}{2}\right)\right)$

As, ${\cos }^{-1}\left(-\theta \right)=\left(\pi -{\cos }^{-1}\theta \right)$

  $\begin{array}{c}\tan \left(\pi -{\cos }^{-1}\pi \left(\frac{1}{2}\right)\right)=-\tan \left(-{\cos }^{-1}\left(\frac{1}{2}\right)\right)\\ =-\left(-\tan \left({\cos }^{-1}\left(\frac{1}{2}\right)\right)\right)\\ =\tan \left({\cos }^{-1}\left(\frac{1}{2}\right)\right)\end{array}$

Recall that tangent function and cosine function related as,

  $\tan \theta =\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }$

Using it,

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

Observe that the value $\frac{1}{2}$ is lies between $-1$ and $1$ .

Recall that for the cosine function and its inverse, the following holds:

  $\begin{array}{c}f\left(f^{-1}\left(x\right)\right)=\cos \left({\cos }^{-1}\left(x\right)\right)\\ =x\end{array}$

Where, $-1\le x\le 1$

Simplify further,

  $\begin{array}{c}\frac{\sqrt{1-{\left[\cos \left({\cos }^{-1}\left(\frac{1}{2}\right)\right)\right]}^2}}{\cos \left({\cos }^{-1}\left(\frac{1}{2}\right)\right)}=\frac{\sqrt{1-{\left(\frac{1}{2}\right)}^2}}{\left(\frac{1}{2}\right)}\\ =\sqrt{3}\end{array}$

Thus, the simplified value of the expression $\tan \left({\cos }^{-1}\left(-\frac{1}{2}\right)\right)$ is $\sqrt{3}$ .