Chapter 3.6, Problem 7E

To determine

To calculate: The value of trigonometric expression $\sin \left(2{\tan }^{-1}\sqrt{2}\right)$ .

Answer to Problem 7E

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

Explanation of Solution

Given information:

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

Formula used:

The inverse trigonometric tangent function is expressed in terms of inverse trigonometric sine function as $2{\tan }^{-1}x=\pi -{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ if $x>1$ .

Calculation:

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

Recall that the inverse trigonometric tangent function is expressed in terms of inverse trigonometric sine function as $2{\tan }^{-1}x=\pi -{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ if $x>1$ .

Apply it,

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

Now substitute the value of $2{\tan }^{-1}\sqrt{2}$ in the provided expression. Use the rule that $\sin \left(\pi -x\right)=-\sin x$ .

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

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