Chapter 14, Problem 7CLR

To determine

To find: The exact values of the six trigonometric functions of $-\frac{13\pi }{6}$ .

Answer to Problem 7CLR

The exact values of the six trigonometric functions of $-\frac{13\pi }{6}$ are $\sin \left(-\frac{13\pi }{6}\right)=-\frac{1}{2}$ , $\cos \left(-\frac{13\pi }{6}\right)=\frac{\sqrt{3}}{2}$ , $\tan \left(-\frac{13\pi }{6}\right)=-\frac{\sqrt{3}}{3}$ , $\csc \left(-\frac{13\pi }{6}\right)=-2$ , $\sec \left(-\frac{13\pi }{6}\right)=\frac{2\sqrt{3}}{3}$ , and $\cot \left(-\frac{13\pi }{6}\right)=-\sqrt{3}$ .

Explanation of Solution

Given:

The angle is $-\frac{13\pi }{6}$ .

Calculation:

It is known that $-\frac{13\pi }{6}$ is co-terminal with $-\frac{\pi }{6}$ or $-30°$ so find the six trigonometric functions corresponding to $-30°$ as follows.

  $\begin{array}{c}\sin \left(-\frac{13\pi }{6}\right)=-\frac{1}{2}\\ \cos \left(-\frac{13\pi }{6}\right)=\frac{\sqrt{3}}{2}\\ \tan \left(-\frac{13\pi }{6}\right)=-\frac{\sqrt{3}}{3}\\ \csc \left(-\frac{13\pi }{6}\right)=-2\end{array}$

And,

  $\begin{array}{c}\sec \left(-\frac{13\pi }{6}\right)=\frac{2\sqrt{3}}{3}\\ \cot \left(-\frac{13\pi }{6}\right)=-\sqrt{3}\end{array}$

Therefore, the exact values of the six trigonometric functions of $-\frac{13\pi }{6}$ are $\sin \left(-\frac{13\pi }{6}\right)=-\frac{1}{2}$ , $\cos \left(-\frac{13\pi }{6}\right)=\frac{\sqrt{3}}{2}$ , $\tan \left(-\frac{13\pi }{6}\right)=-\frac{\sqrt{3}}{3}$ , $\csc \left(-\frac{13\pi }{6}\right)=-2$ , $\sec \left(-\frac{13\pi }{6}\right)=\frac{2\sqrt{3}}{3}$ , and $\cot \left(-\frac{13\pi }{6}\right)=-\sqrt{3}$ .