Chapter 5, Problem 3T

(a)

To determine

To find: The exact value for $\sin \frac{7\pi }{6}$.

Answer to Problem 3T

The exact value of $\sin \frac{7\pi }{6}$ is $-\frac{1}{2}$.

Explanation of Solution

Definition used:

The reference number $\overline{t}$ associated with $t$ is the shortest distance between terminal point determined by $t$ and the $x-$ axis along the unit circle.

If t belongs to first quadrant, reference number is t itself.

If t belongs to second quadrant, reference number is $\pi -t$.

If t belongs to third quadrant, reference number is $t-\pi$.

If t belongs to forth quadrant, reference number is $2\pi -t$.

Calculation:

The terminal point of the trigonometric function is $\frac{7\pi }{6}$.

Since the terminal point is in quadrant III, the reference number is,

$\frac{7\pi }{6}-\pi =\frac{\pi }{6}$

The sin function is negative in third quadrant.

$\begin{array}{c}\sin \left(\frac{7\pi }{6}\right)=-\sin \left(\frac{\pi }{6}\right)\\ \sin \left(\frac{7\pi }{6}\right)=-\frac{1}{2}\left[\sin \frac{\pi }{6}=\frac{1}{2}\right]\end{array}$

Hence, the exact value of $\sin \frac{7\pi }{6}$ is $-\frac{1}{2}$

(b)

To determine

To find: The exact value for $\cos \frac{13\pi }{4}$.

Answer to Problem 3T

The exact value of $\cos \frac{13\pi }{4}$ is $-\frac{\sqrt{2}}{2}$.

Explanation of Solution

The terminal point of the trigonometric function is $\frac{13\pi }{4}$.

Since the terminal point is in quadrant III, the reference number of $\frac{13\pi }{4}$ is $\frac{\pi }{4}$.

The cosine function is negative in third quadrant.

$\begin{array}{c}\cos \left(\frac{13\pi }{4}\right)=-\cos \left(\frac{\pi }{4}\right)\\ \cos \left(\frac{13\pi }{4}\right)=-\frac{1}{\sqrt{2}}\left[\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right]\\ \cos \left(\frac{13\pi }{4}\right)=-\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ \cos \left(\frac{13\pi }{4}\right)=-\frac{\sqrt{2}}{2}\end{array}$

Hence, the exact value of $\cos \frac{13\pi }{4}$ is $-\frac{\sqrt{2}}{2}$

(c)

To determine

To find: The exact value for $\tan \left(-\frac{5\pi }{3}\right)$.

Answer to Problem 3T

The exact value of $\tan \left(-\frac{5\pi }{3}\right)$ is $\sqrt{3}$.

Explanation of Solution

The terminal point of the trigonometric function is $-\frac{5\pi }{3}$.

The negative sign shows angle of rotation is in anti-clockwise direction.

The reference number of $-\frac{5\pi }{3}$ is $\frac{\pi }{3}$.

Since the terminal point is in quadrant I, therefore tan function is negative.

$\begin{array}{l}\tan \left(-\frac{5\pi }{3}\right)=\tan \left(\frac{\pi }{3}\right)\\ \tan \left(-\frac{5\pi }{3}\right)=\sqrt{3}\left[\tan \frac{\pi }{3}=\sqrt{3}\right]\end{array}$

Hence, the exact value of $\tan \left(-\frac{5\pi }{3}\right)$ is $\sqrt{3}$.

(d)

To determine

To find: The exact value for $\csc \frac{3\pi }{2}$.

Answer to Problem 3T

The exact value of $\csc \left(\frac{3\pi }{2}\right)$ is $-1$.

Explanation of Solution

The terminal point of the trigonometric function is $\frac{3\pi }{2}$.

The reference number of $\frac{3\pi }{2}$ is $\frac{\pi }{2}$.

Since the terminal point is on negative y-axis, therefore csc function is negative.

$\begin{array}{c}\csc \left(\frac{3\pi }{2}\right)=\csc \left(\frac{\pi }{2}\right)\\ \csc \left(\frac{3\pi }{2}\right)=\frac{1}{\sin \left(\frac{\pi }{2}\right)}\left[\sin \theta =\frac{1}{\csc \theta }\right]\\ \csc \left(\frac{3\pi }{2}\right)=\frac{1}{-1}\left[\sin \frac{\pi }{2}=1\right]\\ \csc \left(\frac{3\pi }{2}\right)=-1\end{array}$

Hence, the exact value of $\csc \left(\frac{3\pi }{2}\right)$ is $-1$.