Chapter 5.1, Problem 44E

a.

To determine

Reference number for value of t.

Answer to Problem 44E

  ${\theta }_{ref}=\frac{\pi }{6}$

Explanation of Solution

Given information:

  $t=\frac{-7\pi }{6}$

Formula used:

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

  $t=\frac{-7\pi }{6}$

Finding co-terminal angle of $\frac{-7\pi }{6}$

  $\begin{array}{l}2\pi -\frac{7\pi }{6}\\ \frac{12\pi }{6}-\frac{7\pi }{6}\\ \frac{5\pi }{6}\end{array}$

As the angle lies in 2^nd quadrant. So,

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

b.

To determine

Terminal point determined at t.

Answer to Problem 44E

  $P\left(\frac{-\sqrt{3}}{2},\frac{1}{2}\right)$

Explanation of Solution

Given information:

  $t=\frac{-7\pi }{6}$

  $t=\frac{-7\pi }{6}$

As we know co-terminal angle of $\frac{-7\pi }{6}$ is $\frac{5\pi }{6}$ .

  

Above figure shows different terminal points at different angle(t)

For, $\frac{5\pi }{6}$ terminal point is $\left(\frac{-\sqrt{3}}{2},\frac{1}{2}\right)$ .

Therefore, $\frac{-7\pi }{6}$ terminal point is $\left(\frac{-\sqrt{3}}{2},\frac{1}{2}\right)$