Chapter 13.2, Problem 6OE

To determine

To find: The exact values of the trigonometric functions of the number $s$ .

Answer to Problem 6OE

The values of trigonometric functions are

  $\sin s=-\frac{12}{13},\cos s=-\frac{5}{13},\tan s=\frac{12}{5},\cot s=\frac{5}{12},\sec s=-\frac{13}{5},\cos ecs=-\frac{13}{12}$ .

Explanation of Solution

Given information:

Each circle is a unit circle centered at the origin .

  

Calculation:

This is a unit circle ,the angle is ( in radians ) that the line from the origin to the point make with the $x-axis$ is equal to the arc length $s$

Therefore , $\sin s$ is the $y-$ co-ordinate of the given point

  $\sin s=-\frac{12}{13}$

And $\cos s$ is the $y-$ co-ordinate of the given point

  $\cos s=-\frac{5}{13}$

Calculate the other four values

  $\tan s$

  $\begin{array}{c}\tan s=\frac{\sin s}{\cos s}\\ \tan s=\frac{-\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}\\ \tan s=\frac{12}{5}\end{array}$

  $\cot s$

  $\begin{array}{l}\cot s=\frac{\cos s}{\sin s}\\ \cot s=\frac{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}{-\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}\\ \cot s=\frac{5}{12}\end{array}$

  $\sec s$

  $\begin{array}{l}\sec s=\frac{1}{\cos s}\\ \sec s=\frac{1}{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}\\ \sec s=-\frac{13}{5}\end{array}$

  $\cos ecs$

  $\begin{array}{l}\cos ecs=\frac{1}{\sin s}\\ \cos ecs=\frac{1}{-\raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}\\ \cos ecs=-\frac{13}{12}\end{array}$

Therefore, The values of trigonometric functions are

  $\sin s=-\frac{12}{13},\cos s=-\frac{5}{13},\tan s=\frac{12}{5},\cot s=\frac{5}{12},\sec s=-\frac{13}{5},\cos ecs=-\frac{13}{12}$ .