Chapter 7, Problem 25RE

To determine

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

Answer to Problem 25RE

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

Explanation of Solution

Given information:

The expression is $\tan \left({\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$ .

Formula used:

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

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

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

Tangent function and sine function related as,

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

Calculation:

Consider the expression $\tan \left({\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)$ .

Recall that tangent function and sine function related as,

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

Using it,

  $\tan \left({\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)=\frac{\sin \left({\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)}{\sqrt{1-{\left[\sin \left({\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right)\right]}^2}}$

Observe that the value $-\frac{\sqrt{3}}{2}$ is lies between $-1$ and $1$ .

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

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

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

Simplify further,

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

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