Chapter 13.2, Problem 15WE

To determine

To find: The exact values of six trigonometric functions and declare undefined if so.

Answer to Problem 15WE

The value of six trigonometric function are $\sin \left(\frac{8\pi }{2}\right)=\frac{\sqrt{3}}{2}$ , $\cos \left(\frac{8\pi }{2}\right)=\frac{-1}{2}$ , $\tan \left(\frac{8\pi }{2}\right)=-\sqrt{3}$ , $\cot \left(\frac{8\pi }{2}\right)=-\frac{\sqrt{3}}{3}$ , $\sec \left(\frac{8\pi }{2}\right)=-2$ , $\csc \left(\frac{8\pi }{2}\right)=\frac{2\sqrt{3}}{3}$ .

Explanation of Solution

Given information:

The given value of angle is $\frac{8\pi }{3}$ .

Calculation:

The value of angle is $\frac{8\pi }{3}=480°$ . It lies in second quadrant.

The trigonometric table is given below.

$\theta$	$\sin \theta$	$\cos \theta$	$\tan \theta$	$\csc \theta$	$\sec \theta$	$\cot \theta$
$30°$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$	$2$	$2\frac{\sqrt{3}}{3}$	$\sqrt{3}$
$45°$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$	$\sqrt{2}$	$\sqrt{2}$	$1$
$60°$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$	$\frac{2\sqrt{3}}{3}$	$2$	$\frac{\sqrt{3}}{3}$

The value of all trigonometric functions except sine and cosec is negative in this quadrant.

Calculate the values of trigonometric functions.

  $\begin{array}{l}\sin \left(\frac{8\pi }{2}\right)=\frac{\sqrt{3}}{2}\\ \cos \left(\frac{8\pi }{2}\right)=\frac{-1}{2}\\ \tan \left(\frac{8\pi }{2}\right)=-\sqrt{3}\\ \cot \left(\frac{8\pi }{2}\right)=-\frac{\sqrt{3}}{3}\end{array}$

  $\begin{array}{l}\sec \left(\frac{8\pi }{2}\right)=-2\\ \csc \left(\frac{8\pi }{2}\right)=\frac{2\sqrt{3}}{3}\end{array}$

Therefore, the value of six trigonometric function are $\sin \left(\frac{8\pi }{2}\right)=\frac{\sqrt{3}}{2}$ , $\cos \left(\frac{8\pi }{2}\right)=\frac{-1}{2}$ , $\tan \left(\frac{8\pi }{2}\right)=-\sqrt{3}$ , $\cot \left(\frac{8\pi }{2}\right)=-\frac{\sqrt{3}}{3}$ , $\sec \left(\frac{8\pi }{2}\right)=-2$ , $\csc \left(\frac{8\pi }{2}\right)=\frac{2\sqrt{3}}{3}$ .