Chapter 1.6, Problem 40E

To determine

To find: The values of six trigonometric functions.

Answer to Problem 40E

The values of six trigonometry functions are $\sin \theta =\frac{1}{\sqrt{2}}$ , $\cos \theta =-\frac{1}{\sqrt{2}}$ , $\tan \theta =-1$ , $\cot \theta =-1$ , $\sec \theta =-\sqrt{2}$ and $\csc \theta =\sqrt{2}$ .

Explanation of Solution

Given information: The given point is $P\left(-2,2\right)$ .

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:

The value of x is $-2$ and y is 2. By the terminal point the value of $r$ can be calculated as:

  $\begin{array}{c}r=\sqrt{{\left(-2\right)}^2+2^2}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

The values of six trigonometric ratios are:

  $\begin{array}{c}\sin \theta =\frac{1}{\sqrt{2}}\\ \cos \theta =-\frac{1}{\sqrt{2}}\\ \tan \theta =-1\\ \cot \theta =-1\\ \sec \theta =-\sqrt{2}\\ \csc \theta =\sqrt{2}\end{array}$

Therefore, the values of six trigonometry functions are $\sin \theta =\frac{1}{\sqrt{2}}$ , $\cos \theta =-\frac{1}{\sqrt{2}}$ , $\tan \theta =-1$ , $\cot \theta =-1$ , $\sec \theta =-\sqrt{2}$ and $\csc \theta =\sqrt{2}$ .