Chapter 5.1, Problem 45E

a.

To determine

Reference number for value of t.

Answer to Problem 45E

  ${\theta }_{ref}=\frac{\pi }{4}$

Explanation of Solution

Given information:

  $t=\frac{13\pi }{4}$

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

  $t=\frac{13\pi }{4}$

Finding co-terminal angle of $\frac{13\pi }{4}$

  $\begin{array}{l}\frac{13\pi }{4}-3\pi \\ \frac{13\pi }{4}-\frac{12\pi }{4}\\ \frac{\pi }{4}\end{array}$

As the angle lies in 1^st quadrant. So,

  $\begin{array}{l}{\theta }_{ref}=\theta \\ {\theta }_{ref}=\frac{\pi }{4}\end{array}$

b.

To determine

Terminal point determined at t.

Answer to Problem 45E

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

Explanation of Solution

Given information:

  $t=\frac{13\pi }{4}$

  $t=\frac{13\pi }{4}$

As we know co-terminal angle of $\frac{13\pi }{4}$ is $\frac{\pi }{4}$ .

  

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

For, $\frac{\pi }{4}$ terminal point is $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ .

Therefore, $\frac{13\pi }{4}$ terminal point is $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$