Chapter 5.2, Problem 6E

a.

To determine

Exact value Trigonometric function at the given real number.

Answer to Problem 6E

  $\sin \frac{5\pi }{6}=\frac{1}{2}$

Explanation of Solution

Given information:

  $\sin \frac{5\pi }{6}$

Formula used:

Finding value by using reference angle of given angle.

Reference angle for different quadrant:

1^st quadrant: ${\theta }_{ref}=\theta$

2^nd quadrant: ${\theta }_{ref}=\pi -\theta$

3^rd quadrant: ${\theta }_{ref}=\theta -\pi$

4^th quadrant: ${\theta }_{ref}=2\pi -\theta$

As angle lies in 2^nd quadrant. So,

  $\begin{array}{l}{\theta }_{ref}=\pi -\theta \\ {\theta }_{ref}=\pi -\frac{5\pi }{6}\\ {\theta }_{ref}=\frac{6\pi }{6}-\frac{5\pi }{6}\\ {\theta }_{ref}=\frac{\pi }{6}\end{array}$

As $\sin \theta$ is positive in 2^nd quadrant, So

  $\sin \frac{5\pi }{6}=\sin \frac{\pi }{6}=\frac{1}{2}$

b.

To determine

Exact value Trigonometric function at the given real number.

Answer to Problem 6E

  $\cos \frac{5\pi }{6}=\frac{-\sqrt{3}}{2}$

Explanation of Solution

Given information:

  $\cos \frac{5\pi }{6}$

Formula used:

Finding value by using reference angle of given angle.

Reference angle for different quadrant:

1^st quadrant: ${\theta }_{ref}=\theta$

2^nd quadrant: ${\theta }_{ref}=\pi -\theta$

3^rd quadrant: ${\theta }_{ref}=\theta -\pi$

4^th quadrant: ${\theta }_{ref}=2\pi -\theta$

As angle lies in 2^nd quadrant. So,

  $\begin{array}{l}{\theta }_{ref}=\pi -\theta \\ {\theta }_{ref}=\pi -\frac{5\pi }{6}\\ {\theta }_{ref}=\frac{6\pi }{6}-\frac{5\pi }{6}\\ {\theta }_{ref}=\frac{\pi }{6}\end{array}$

As $\cos \theta$ is negative in 2^nd quadrant, So

  $\cos \frac{5\pi }{6}=\cos \frac{\pi }{6}=-\frac{\sqrt{3}}{2}$

c.

To determine

Exact value Trigonometric function at the given real number.

Answer to Problem 6E

  $\tan \frac{5\pi }{6}=-\frac{1}{\sqrt{3}}$

Explanation of Solution

Given information:

  $\tan \frac{5\pi }{6}$

Formula used:

Finding value by using reference angle of given angle.

Reference angle for different quadrant:

1^st quadrant: ${\theta }_{ref}=\theta$

2^nd quadrant: ${\theta }_{ref}=\pi -\theta$

3^rd quadrant: ${\theta }_{ref}=\theta -\pi$

4^th quadrant: ${\theta }_{ref}=2\pi -\theta$

As angle lies in 2^nd quadrant. So,

  $\begin{array}{l}{\theta }_{ref}=\pi -\theta \\ {\theta }_{ref}=\pi -\frac{5\pi }{6}\\ {\theta }_{ref}=\frac{6\pi }{6}-\frac{5\pi }{6}\\ {\theta }_{ref}=\frac{\pi }{6}\end{array}$

As $\tan \theta$ is negative in 2^nd quadrant, So

  $\tan \frac{\pi }{6}=\tan \frac{\pi }{6}=-\frac{1}{\sqrt{3}}$