Chapter 5.3, Problem 33E

To determine

To find: The values of six trigonometric functions for $\text{angle }\theta$ in standard position if a point with the coordinates $\left(1,-8\right)$ lies on its terminal side.

Answer to Problem 33E

  $\begin{array}{l}\sin \theta =\frac{-8\sqrt{65}}{65},\csc \theta =\frac{-\sqrt{65}}{8}\\ \cos \theta =\frac{\sqrt{65}}{65},\sec \theta =\sqrt{65}\\ \tan \theta =-8,\cot \theta =\frac{-1}{8}\end{array}$

Explanation of Solution

Given information: Coordinates $\left(1,-8\right)$

Calculation:

  $\text{hyp = }\sqrt{65}\text{, opp = }-8\text{ and adj = 1}$

  $\begin{array}{l}\sin \theta =\frac{-8}{\sqrt{65}}=\frac{-8\sqrt{65}}{65}\\ \csc \theta =\frac{1}{\frac{-8\sqrt{65}}{65}}=\frac{-65}{8\sqrt{65}}=\frac{-65\sqrt{65}}{8\times 65}=\frac{-\sqrt{65}}{8}\\ \cos \theta =\frac{1}{\sqrt{65}}=\frac{\sqrt{65}}{65}\\ \sec \theta =\frac{1}{\cos }=\frac{1}{\frac{\sqrt{65}}{65}}=\frac{65}{\sqrt{65}}=\sqrt{65}\\ \tan \theta =\frac{-8}{1}=-8\\ \cot \theta =\frac{1}{\tan }=\frac{1}{-8}=\frac{-1}{8}\end{array}$