Chapter 14, Problem 2CLR

To determine

To find: The values of the six trigonometric functions of an angle $\theta$ in standard position whose terminal side passes through the point $\left(-2,5\right)$ .

Answer to Problem 2CLR

The six trigonometric function values are $\sin \theta =\frac{5}{\sqrt{29}}$ , $\cos \theta =-\frac{2}{\sqrt{29}}$ , $\tan \theta =-\frac{5}{2}$ , $\csc \theta =\frac{\sqrt{29}}{3}$ , $\sec \theta =-\frac{\sqrt{29}}{2}$ , and $\cot \theta =-\frac{2}{5}$ .

Explanation of Solution

Given:

The terminal point is $\left(-2,5\right)$ .

Calculation:

Consider the following figure.

  

Determine the value of $r$ .

  $\begin{array}{c}r=\sqrt{x^2+y^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(5\right)}^2}\\ =\sqrt{29}\end{array}$

The six trigonometric function values for the given angle are as follows.

  $\begin{array}{c}\sin \theta =\frac{5}{\sqrt{29}}\\ \cos \theta =-\frac{2}{\sqrt{29}}\\ \tan \theta =-\frac{5}{2}\end{array}$

By using the reciprocal of the trigonometric functions the other trigonometric function values are as follows.

  $\begin{array}{c}\csc \theta =\frac{\sqrt{29}}{3}\\ \sec \theta =-\frac{\sqrt{29}}{2}\\ \cot \theta =-\frac{2}{5}\end{array}$

Therefore, the six trigonometric function values are $\sin \theta =\frac{5}{\sqrt{29}}$ , $\cos \theta =-\frac{2}{\sqrt{29}}$ , $\tan \theta =-\frac{5}{2}$ , $\csc \theta =\frac{\sqrt{29}}{3}$ , $\sec \theta =-\frac{\sqrt{29}}{2}$ , and $\cot \theta =-\frac{2}{5}$ .