Chapter 7.4, Problem 25E

To determine

To find: the value of $\sin 2\theta ,\cos 2\theta$ and $\tan 2\theta$ by using given information.

Answer to Problem 25E

  $\begin{array}{l}\sin 2\theta =\frac{12}{13}\end{array}$

  $\begin{array}{c}\cos 2\theta =\frac{5}{13}\end{array}$

  $\begin{array}{c}\tan 2\theta =\frac{12}{5}\end{array}$

Explanation of Solution

Given information:

  $\cot \theta =\frac{3}{2},\text{ 180}°<\theta <270°$

Calculation:

Evaluate $\sin \theta$ as follows:

  $\begin{array}{c}\csc \theta =\sqrt{1+{\cot }^2\theta }\left[\text{Pythagorean identity}\right]\end{array}$

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

  $\begin{array}{c}\csc \theta =\frac{\sqrt{13}}{2}\end{array}$

  $\begin{array}{c}\sin \theta =\frac{2}{\sqrt{13}}\left[\because \sin \theta =\frac{1}{\csc \theta }\right]\end{array}$

In third quadrant the value of sine is negative. Thus the value of $\sin \theta$ is $-\frac{2}{\sqrt{13}}$ .

Evaluate $\cos \theta$ by using Pythagorean identity as follows:

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

  $\begin{array}{c}=\sqrt{1-{\left(-\frac{2}{\sqrt{13}}\right)}^2}\left[\text{Put the value of sin}\theta \right]\end{array}$

  $\begin{array}{c}=\sqrt{1-\frac{4}{13}}\end{array}$

  $\begin{array}{c}=\frac{3}{\sqrt{13}}\end{array}$

In third quadrant the value of cosine is negative. Thus the value of $\cos \theta$ is $-\frac{3}{\sqrt{13}}$ .

Now evaluate $\sin 2\theta$ by using the double angle identity of sine as follows:

  $\begin{array}{c}\sin 2\theta =2\sin \theta \cos \theta \end{array}$

  $\begin{array}{c}=2\left(-\frac{2}{\sqrt{13}}\right)\left(-\frac{3}{\sqrt{13}}\right)\left[\text{Put the value of sin}\theta \text{ and cos}\theta \right]\end{array}$

  $\begin{array}{c}=\frac{12}{13}\end{array}$

Thus $\sin 2\theta$ is $\frac{4\sqrt{21}}{5}$ .

Now evaluate $\cos 2\theta$ by using the double angle identity of cosine as follows:

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

  $\begin{array}{c}=1-2{\left(-\frac{2}{\sqrt{13}}\right)}^2\left[\text{Put the value of sin}\theta \right]\end{array}$

  $\begin{array}{c}=1-\frac{8}{13}\end{array}$

  $\begin{array}{c}=\frac{5}{13}\end{array}$

Thus $\cos 2\theta$ is $\frac{5}{13}$ .

Now evaluate $\tan 2\theta$ by using the quotient identity of tangent as follows:

  $\begin{array}{c}\tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta }\end{array}$

  $\begin{array}{c}=\frac{\left(\frac{12}{13}\right)}{\left(\frac{5}{13}\right)}\left[\text{Put the value of sin2}\theta \text{ and cos2}\theta \right]\end{array}$

  $\begin{array}{c}=\frac{12}{5}\end{array}$

Thus the value of $\tan 2\theta$ is $\frac{12}{5}$ .