Chapter 7.4, Problem 27E

To determine

To find: the value of $\tan 2\alpha$ using the given information.

Answer to Problem 27E

  $\frac{2\sqrt{14}}{5}$

Explanation of Solution

Given information:

  $\cos \alpha =-\frac{\sqrt{2}}{3}\left[\alpha \text{ is in second quadrant}\right]$

Calculation:

Evaluate $\sin \alpha$ by using Pythagorean identity as follows:

  $\begin{array}{c}\sin \alpha =\sqrt{1-{\cos }^2\alpha }\end{array}$

  $\begin{array}{c}=\sqrt{1-{\left(\raisebox{1ex}{$-\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}^2}\left[\text{Put the value of cos}\alpha \right]\end{array}$

  $\begin{array}{c}=\sqrt{1-\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$9$}\right.}\end{array}$

  $\begin{array}{c}=\frac{\sqrt{7}}{3}\end{array}$

Thus the value of $\sin \alpha$ in second quadrant is $\frac{\sqrt{7}}{3}$ .

Now evaluate $\tan \alpha$ by using the quotient identity of tangent as follows:

  $\begin{array}{c}\tan \alpha =\frac{\sin \alpha }{\cos \alpha }\end{array}$

  $\begin{array}{c}=\frac{\left(\raisebox{1ex}{$\sqrt{7}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}{\left(\raisebox{1ex}{$-\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}\left[\text{Put the value of sin}\alpha \text{ and cos}\alpha \right]\end{array}$

  $\begin{array}{c}=-\frac{\sqrt{7}}{\sqrt{2}}\end{array}$

  $\begin{array}{c}=-\frac{\sqrt{14}}{2}\left[\text{Multiply and divide by }\sqrt{2}\right]\end{array}$

Thus the value of $\tan \alpha$ is $-\frac{\sqrt{14}}{2}$ .

Now evaluate $\tan 2\alpha$ by using the double angle identity of tangent as follows:

  $\begin{array}{c}\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }\end{array}$

  $\begin{array}{c}=\frac{2\left(-\frac{\sqrt{14}}{2}\right)}{1-{\left(-\frac{\sqrt{14}}{2}\right)}^2}\left[\text{Put the value of }\tan \alpha \right]\end{array}$

  $\begin{array}{c}=\frac{-\sqrt{14}}{1-\frac{7}{2}}\end{array}$

  $\begin{array}{c}=\frac{-\sqrt{14}}{\frac{-5}{2}}\end{array}$

  $\begin{array}{c}=\frac{2\sqrt{14}}{5}\end{array}$

Thus $\tan 2\theta$ is $\frac{2\sqrt{14}}{5}$ .