Chapter 1.6, Problem 38E

To determine

To find: The values of six trigonometric functions.

Answer to Problem 38E

The values of trigonometry are $\sin \theta =\frac{-5}{13}$ , $\cos \theta =\frac{12}{13}$ , $\tan \theta =-\frac{5}{12}$ , $\csc \theta =\frac{13}{-5}$ , $\sec \theta =\frac{13}{12}$ , and $\cot \theta =-\frac{12}{5}$ .

Explanation of Solution

Given information: The given value is $\theta ={\tan }^{-1}\left(-\frac{5}{12}\right)$ .

Formula used: The six trigonometric functions with an angle $\theta$ can be calculated as:

  $\sin \theta =\frac{y}{r}$ , $\cos \theta =\frac{x}{r}$ , $\tan \theta =\frac{y}{x}$ , $\cot \theta =\frac{x}{y}$ , $\csc \theta =\frac{r}{y}$ , $\sec \theta =\frac{r}{x}$

Calculation:

Simplify the given value of angle.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-\frac{5}{12}\right)\\ \tan \theta =-\frac{5}{12}\end{array}$

So, $\cot \theta =-\frac{12}{5}$ .

Now, by using Pythagorean Theorem as:

  $\begin{array}{c}r=\sqrt{{\left(-5\right)}^2+{12}^2}\\ =\sqrt{169}\\ =13\end{array}$

Compare the value of $\tan \theta$ with $\tan \theta =\frac{y}{x}$ to get,

  $\begin{array}{c}y=-5\\ x=12\end{array}$

The remaining trigonometric ratios are:

  $\begin{array}{c}\sin \theta =\frac{-5}{13}\\ \csc \theta =\frac{13}{-5}\\ \cos \theta =\frac{12}{13}\\ \sec \theta =\frac{13}{12}\end{array}$

Therefore, the values of trigonometry are $\sin \theta =\frac{-5}{13}$ , $\cos \theta =\frac{12}{13}$ , $\tan \theta =-\frac{5}{12}$ , $\csc \theta =\frac{13}{-5}$ , $\sec \theta =\frac{13}{12}$ , and $\cot \theta =-\frac{12}{5}$ .