Chapter 14, Problem 11CLR

(a)

To determine

To find: The value of $\cos \frac{\varphi }{2}$ .

Answer to Problem 11CLR

The value of $\cos \frac{\varphi }{2}=-\frac{\sqrt{10}}{10}$ .

Explanation of Solution

Given:

It is given $\sin \varphi =-\frac{3}{5}$ and $90°<\varphi <270°$ .

Calculation:

Use the identity ${\sin }^2\varphi +{\cos }^2\varphi =1$ .

Substitute $-\frac{3}{5}$ for $\sin \varphi$ in the equation ${\sin }^2\varphi +{\cos }^2\varphi =1$ to find $\cos \varphi$ .

  $\begin{array}{c}{\left(-\frac{3}{5}\right)}^2+{\cos }^2\varphi =1\\ {\cos }^2\varphi =1-\frac{9}{25}\\ =\frac{16}{25}\\ \cos \varphi =\pm \frac{4}{5}\end{array}$

Since $90°<\varphi <270°$ , $\varphi$ is in the third quadrant and it is known that in the third quadrant $\cos \varphi$ is negative. So, $\cos \varphi =-\frac{4}{5}$ .

Substitute $-\frac{4}{5}$ for $\cos \varphi$ in the equation $\cos \frac{\varphi }{2}=\pm \sqrt{\frac{1+\cos \varphi }{2}}$ .

  $\begin{array}{c}\cos \frac{\varphi }{2}=\pm \sqrt{\frac{1-\frac{4}{5}}{2}}\\ =\pm \sqrt{\frac{1}{10}}\\ =\pm \frac{\sqrt{10}}{10}\end{array}$

Since it is in third quadrant so $\cos \frac{\varphi }{2}=-\frac{\sqrt{10}}{10}$ .

Therefore, the value of $\cos \frac{\varphi }{2}=-\frac{\sqrt{10}}{10}$ .

(b)

To determine

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

Answer to Problem 11CLR

The value of $\cos 2\varphi =\frac{7}{25}$ .

Explanation of Solution

Given:

It is given $\sin \varphi =-\frac{3}{5}$ and $90°<\varphi <270°$ .

Calculation:

Substitute $-\frac{3}{5}$ for $\sin \varphi$ and $-\frac{4}{5}$ for $\cos \varphi$ in the equation $\cos 2\varphi ={\cos }^2\varphi -{\sin }^2\varphi$ to find $\cos 2\varphi$ .

  $\begin{array}{c}\cos 2\varphi ={\left(-\frac{4}{5}\right)}^2-{\left(-\frac{3}{5}\right)}^2\\ =\frac{16}{25}-\frac{9}{25}\\ =\frac{7}{25}\end{array}$

Therefore, the value of $\cos 2\varphi =\frac{7}{25}$ .

(c)

To determine

To find: The value of $\tan \left(180°-\varphi \right)$ .

Answer to Problem 11CLR

The value of $\tan \left(180°-\varphi \right)=-\frac{3}{4}$ .

Explanation of Solution

Given:

It is given $\sin \varphi =-\frac{3}{5}$ and $90°<\varphi <270°$ .

Calculation:

It is known that $\tan \left(180°-\varphi \right)=-\tan \varphi$ .

Use the identity $\tan \varphi =\frac{\sin \varphi }{\cos \varphi }$ .

Substitute $-\frac{3}{5}$ for $\sin \varphi$ and $-\frac{4}{5}$ for $\cos \varphi$ in the equation $\tan \varphi =\frac{\sin \varphi }{\cos \varphi }$ .

  $\begin{array}{c}-\tan \varphi =-\left(\frac{-\frac{3}{5}}{-\frac{4}{5}}\right)\\ =-\frac{3}{4}\end{array}$

Therefore, the value of $\tan \left(180°-\varphi \right)=-\frac{3}{4}$ .

(d)

To determine

To find: The value of $\sin \left(90°+\varphi \right)$ .

Answer to Problem 11CLR

The value of $\sin \left(90°+\varphi \right)=-\frac{4}{5}$ .

Explanation of Solution

Given:

It is given $\sin \varphi =-\frac{3}{5}$ and $90°<\varphi <270°$ .

Calculation:

It is known that $\sin \left(90°+\varphi \right)=\cos \varphi$ .

Substitute $-\frac{4}{5}$ for $\cos \varphi$ in the equation $\sin \left(90°+\varphi \right)=\cos \varphi$ .

  $\sin \left(90°+\varphi \right)=-\frac{4}{5}$

Therefore, the value of $\sin \left(90°+\varphi \right)=-\frac{4}{5}$ .