Chapter 12, Problem 12CR

(a)

To determine

To calculate: The value of $\cos \theta$ if $\sin \theta =\frac{1}{4}$, and $\theta$ is in second quadrant.

Answer to Problem 12CR

Solution:

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

Explanation of Solution

Given information:

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

Formula used:

${\sin }^{2}x+{\cos }^{2}x=1$

Calculation:

By using trigonometric identity,

${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow {\cos }^2\theta =1-{\sin }^2\theta \Rightarrow \cos \theta =\pm \sqrt{1-{\sin }^2\theta }$

As $\theta$ is in second quadrant, implies $\cos \theta <0$,

$\Rightarrow \cos \theta =-\sqrt{1-{\sin }^2\theta }$

Plug in $\sin \theta =\frac{1}{4}$,

$\Rightarrow \cos \theta =-\sqrt{1-{\left(\frac{1}{4}\right)}^2}$

$\Rightarrow \cos \theta =-\sqrt{1-\frac{1}{16}}=-\sqrt{\frac{15}{16}}=-\frac{\sqrt{15}}{4}$

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

(b)

To determine

To calculate: The value of $\tan \theta$ if $\sin \theta =\frac{1}{4}$, and $\theta$ is in second quadrant.

Answer to Problem 12CR

Solution:

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

Explanation of Solution

Given information:

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

Formula used:

$\tan \theta =\frac{\sin \theta }{\cos \theta }$

Calculation:

From part (a), $\cos \theta =-\frac{\sqrt{15}}{4}$.

By using formula,

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\left(\frac{1}{4}\right)}{\left(-\frac{\sqrt{15}}{4}\right)}=\left(\frac{1}{4}\right)\left(-\frac{4}{\sqrt{15}}\right)$

$\Rightarrow \tan \theta =-\frac{1}{\sqrt{15}}\cdot \frac{\sqrt{15}}{\sqrt{15}}=-\frac{\sqrt{15}}{15}$

Hence, the value of $\tan \theta$ is $-\frac{\sqrt{15}}{15}$.

(c)

To determine

To calculate: The value of $\sin \left(2\theta \right)$ if $\sin \theta =\frac{1}{4}$, and $\theta$ is in second quadrant.

Answer to Problem 12CR

Solution:

The value of $\sin \left(2\theta \right)$ is $-\frac{\sqrt{15}}{8}$.

Explanation of Solution

Given information:

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

Formula used:

$\sin \left(2\theta \right)=2\sin \theta \cos \theta$

Calculation:

From part (a), $\cos \theta =-\frac{\sqrt{15}}{4}$.

By using formula,

$\Rightarrow \sin \left(2\theta \right)=2\left(\frac{1}{4}\right)\left(-\frac{\sqrt{15}}{4}\right)=-\frac{\sqrt{15}}{8}$

Hence, the value of $\sin \left(2\theta \right)$ is $-\frac{\sqrt{15}}{8}$.

(d)

To determine

To calculate: The value of $\cos \left(2\theta \right)$ if $\sin \theta =\frac{1}{4}$, and $\theta$ is in second quadrant.

Answer to Problem 12CR

Solution:

The value of $\cos \left(2\theta \right)$ is $\frac{7}{8}$.

Explanation of Solution

Given information:

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

Formula used:

$\cos \left(2\theta \right)={\cos }^2\theta -{\sin }^2\theta$

Calculation:

From part (a), $\cos \theta =-\frac{\sqrt{15}}{4}$.

By using formula,

$\Rightarrow \cos \left(2\theta \right)={\left(-\frac{\sqrt{15}}{4}\right)}^2-{\left(\frac{1}{4}\right)}^2$

$\Rightarrow \cos \left(2\theta \right)=\frac{15}{16}-\frac{1}{16}=\frac{14}{16}=\frac{7}{8}$

Hence, the value of $\cos \left(2\theta \right)$ is $\frac{7}{8}$.

(e)

To determine

To calculate: The value of $\sin \left(\frac{1}{2}\theta \right)$ if $\sin \theta =\frac{1}{4}$, and $\theta$ is in second quadrant.

Answer to Problem 12CR

Solution:

The value of $\sin \left(\frac{1}{2}\theta \right)$ is $\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}$.

Explanation of Solution

Given information:

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

Formula used:

$\sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{1-\cos \theta }{2}}$

Calculation:

Given that $\frac{\pi }{2}<\theta <\pi \Rightarrow \frac{\pi }{4}<\frac{\theta }{2}<\frac{\pi }{2}$.

$\Rightarrow \frac{\theta }{2}$ is in Quadrant I $\Rightarrow \sin \left(\frac{\theta }{2}\right)>0$.

From part (a), $\cos \theta =-\frac{\sqrt{15}}{4}$.

By using formula,

$\sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{1-\cos \theta }{2}}=\sqrt{\frac{1-\left(-\frac{\sqrt{15}}{4}\right)}{2}}$.

$\Rightarrow \sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{\left(\frac{4+\sqrt{15}}{4}\right)}{2}}=\sqrt{\frac{4+\sqrt{15}}{8}}$.

$\Rightarrow \sin \left(\frac{\theta }{2}\right)=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}$.

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