Chapter 12, Problem 12CR

(a)

To determine

To find: The value of $\cos \theta$ , if $\sin \theta =\frac{1}{4}$ .

Answer to Problem 12CR

The value of $\cos \theta =-\frac{\sqrt{15}}{4}$ .

Explanation of Solution

Given information:

The given, $\sin \theta =\frac{1}{4}$ , and $\theta$ is in second quadrant.

Calculation:

Calculate the $\theta$ .

  $\begin{array}{c}\sin \theta =\frac{1}{4}\\ \theta ={\sin }^{-1}\left(\frac{1}{4}\right)\\ \theta ={14.4475}^o\end{array}$

  $\theta$ is in second quadrant.

  $\begin{array}{c}\sin \left(180-\theta \right)=\frac{1}{4}\\ 180-\theta ={\sin }^{-1}\left(\frac{1}{4}\right)\\ 180-\theta ={14.4775}^o\\ \theta ={165.5225}^o\end{array}$

  $\begin{array}{c}\cos \theta =\sqrt{1-{\sin }^2\theta }\\ =\sqrt{1-{\left(\frac{1}{4}\right)}^2}\\ =\sqrt{1-\frac{1}{16}}\\ =\sqrt{\frac{15}{16}}\\ \cos \theta =\frac{\sqrt{15}}{4}\end{array}$

  $\theta$ is in second quadrant.

Therefore, the value of $\cos \theta =-\frac{\sqrt{15}}{4}$ .

(b)

To determine

To find: The value of $\tan \theta$ , if $\sin \theta =\frac{1}{4}$ .

Answer to Problem 12CR

The value of $\tan \theta =-\frac{1}{\sqrt{15}}$ .

Explanation of Solution

Given information:

The given, $\sin \theta =\frac{1}{4}$ , and $\theta$ is in second quadrant.

Calculation:

Calculate the $\tan \theta$ .

  $\begin{array}{c}\tan \theta =\frac{\sin \theta }{\cos \theta }\\ \tan \theta =\frac{\frac{1}{4}}{-\frac{\sqrt{15}}{4}}\\ =-\frac{1}{4}\times \frac{4}{\sqrt{15}}\\ =-\frac{1}{\sqrt{15}}\end{array}$

Therefore, the value of $\tan \theta =-\frac{1}{\sqrt{15}}$

(c)

To determine

To find: The value of $\sin 2\theta$ , if $\sin \theta =\frac{1}{4}$ .

Answer to Problem 12CR

The value of $\sin 2\theta =-\frac{\sqrt{15}}{8}$ .

Explanation of Solution

Given information:

The given, $\sin \theta =\frac{1}{4}$ , and $\theta$ is in second quadrant.

Calculation:

Calculate the $\tan \theta$ .

  $\begin{array}{c}\sin 2\theta =2\sin \theta \cos \theta \\ =2\times \frac{1}{4}\times \frac{\sqrt{15}}{4}\\ =\frac{\sqrt{15}}{4}\end{array}$

Therefore, the value of $\sin 2\theta =-\frac{\sqrt{15}}{8}$ .

(d)

To determine

To find: The value of $\cos 2\theta$ , if $\sin \theta =\frac{1}{4}$ .

Answer to Problem 12CR

The value of $\cos 2\theta =\frac{7}{8}$ .

Explanation of Solution

Given information:

The given $\sin \theta =\frac{1}{4}$ , and $\theta$ is in second quadrant.

Calculation:

Calculate the $\tan \theta$ .

  $\begin{array}{c}\cos 2\theta =\sqrt{1-{\sin }^{2}2\theta }\\ =\sqrt{1-{\left(\frac{\sqrt{15}}{8}\right)}^2}\\ =\sqrt{1-\frac{15}{64}}\\ =\sqrt{\frac{49}{64}}\\ \cos 2\theta =\frac{7}{8}\end{array}$

Therefore, The value of $\cos 2\theta =\frac{7}{8}$ .

(e)

To determine

To find: The value of $\sin \frac{\theta }{2}$ , if $\sin \theta =\frac{1}{4}$ .

Answer to Problem 12CR

  $\sin \frac{\theta }{2}=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}$ .

Explanation of Solution

Given information:

The given $\sin \theta =\frac{1}{4}$ , and $\theta$ is in second quadrant.

Calculation:

Calculate the $\tan \theta$ .

  $\begin{array}{c}{\sin }^2\left(\frac{\theta }{2}\right)=\frac{1-\cos \theta }{2}\\ {\sin }^{}\left(\frac{\theta }{2}\right)=\sqrt{\frac{1-\cos \theta }{2}}\\ =\sqrt{\frac{1-\left(-\frac{\sqrt{15}}{4}\right)}{2}}\\ =\sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}}\\ =\sqrt{\frac{\frac{4+\sqrt{15}}{4}}{2}}\\ =\sqrt{\frac{4+\sqrt{15}}{8}}\\ \sin \left(\frac{\theta }{2}\right)=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}\end{array}$

Therefore, the value of $\sin \left(\frac{\theta }{2}\right)=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}$ .