Chapter 13.9, Problem 5ST

(a)

To determine

To find: The value of $\cos 2\theta$.

Answer to Problem 5ST

The $\cos 2\theta =\frac{41}{841}$ .

Explanation of Solution

Given information:

The given information is ${180}^{\circ }<\theta <{360}^{\circ }$ and $\cos \theta =-\frac{21}{29}$ .

Calculation:

We know that $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$ .

Now, we have to find ${\cos }^2\theta$ and ${\sin }^2\theta$ .

  $\begin{array}{l}{\cos }^2\theta ={\left(-\frac{21}{29}\right)}^2\\ {\cos }^2\theta =\left(\frac{441}{841}\right)\end{array}$

We know that ${\sin }^2\theta =1-{\cos }^2\theta$ .

  $\begin{array}{l}{\sin }^2\theta =1-{\cos }^2\theta \\ {\sin }^2\theta =1-\frac{441}{841}\\ {\sin }^2\theta =\frac{400}{841}\end{array}$

Now,

  $\begin{array}{l}\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta \\ \cos 2\theta =\frac{441}{841}-\frac{400}{841}\\ \cos 2\theta =\frac{41}{841}\end{array}$

Therefore, the $\cos 2\theta =\frac{41}{841}$ .

(b)

To determine

To find: The value of $\sin 2\theta$.

Answer to Problem 5ST

The $\sin 2\theta =\frac{840}{841}$ .

Explanation of Solution

Given information:

The given information is ${180}^{\circ }<\theta <{360}^{\circ }$ and $\cos \theta =-\frac{21}{29}$ .

Calculation:

We know that $\sin 2\theta =2\sin \theta \cos \theta$ .

Now, we have to find $\sin \theta$ .

We know that ${\sin }^2\theta =\frac{400}{841}$ .

  $\begin{array}{l}{\sin }^2\theta =\frac{400}{841}\\ \sin \theta =\sqrt{\frac{400}{841}}\\ \sin \theta =\pm \frac{20}{29}\end{array}$

Since $\sin \theta$ is negative in the third and fourth quadrant, we can take value as $\sin \theta =-\frac{20}{29}$

  $\begin{array}{l}\sin 2\theta =2\sin \theta \cos \theta \\ \sin 2\theta =2\left(-\frac{20}{29}\right)\left(-\frac{21}{29}\right)\\ \sin 2\theta =\frac{840}{841}\end{array}$

Therefore, the $\sin 2\theta =\frac{840}{841}$ .

(c)

To determine

To find: The value of $\cos \left(\frac{\theta }{2}\right)$.

Answer to Problem 5ST

The $\cos \left(\frac{\theta }{2}\right)=-2\frac{\sqrt{29}}{29}$ .

Explanation of Solution

Given information:

The given information is ${180}^{\circ }<\theta <{360}^{\circ }$ and $\cos \theta =-\frac{21}{29}$ .

Calculation:

We know that $\cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\cos \theta }{2}}$ .

Substitute the value of $\cos \theta$ in the equation

  $\begin{array}{l}\cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\cos \theta }{2}}\\ \cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\left(-\frac{21}{29}\right)}{2}}\\ \cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{\frac{29-21}{29}}{2}}\end{array}$

  $\begin{array}{l}\cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{\frac{8}{29}}{2}}\\ \cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{4}{29}}\\ \cos \left(\frac{\theta }{2}\right)=\pm 2\sqrt{\frac{1}{29}}\end{array}$

Multiply the numerator and the denominator by $\sqrt{29}$ .

  $\begin{array}{l}\cos \left(\frac{\theta }{2}\right)=\pm 2\sqrt{\frac{1}{29}}\left(\sqrt{\frac{29}{29}}\right)\\ \cos \left(\frac{\theta }{2}\right)=\pm 2\frac{\sqrt{29}}{29}\end{array}$

From the given interval for we find that $90°<\frac{\theta }{2}<180°$ Sine cosine function is negative in the third quadrant, we need take only the negative value for $\cos \left(\frac{\theta }{2}\right)$ .

Therefore, the $\cos \left(\frac{\theta }{2}\right)=-2\frac{\sqrt{29}}{29}$ .