Chapter 9.3, Problem 4GP

Solution

To evaluate: the value of six trigonometric functions of $\theta =180°$ .

  $\sin \theta =0,\cos \theta =-1,\tan \theta =0,\cot \theta =\text{undefined},\csc \theta =\text{undefined},\sec \theta =-1$

Given information:

The angle is $\theta =180°$ .

Definition Used:

General definitions of trigonometric functions:

  

Let $\theta$ be an angle in standard position, and let $\left(x,y\right)$ be the point where the terminal side of $\theta$ intersects the circle $x^2+y^2=r^2$ . The six trigonometric functions of $\theta$ are defined as follows:

  $\begin{array}{l}\sin \theta =\frac{y}{r}\\ \csc \theta =\frac{r}{y},y\ne 0\\ \cos \theta =\frac{x}{r}\\ \sec \theta =\frac{r}{x},x\ne 0\\ \tan \theta =\frac{y}{x},x\ne 0\\ \cot \theta =\frac{x}{y},y\ne 0\end{array}$

The Unit Circle

The circle $x^2+y^2=1$ , which has center $\left(0,0\right)$ and radius 1, is called unit circle.

  $\begin{array}{c}\sin \theta =\frac{y}{r}\\ =\frac{y}{1}\\ =y\\ \cos \theta =\frac{x}{r}\\ =\frac{x}{1}\\ =x\end{array}$

Explanation:

Draw the unit circle, then draw the angle $\theta =180°$ in standard position. The terminal side of $\theta$ intersects the unit circle at $\left(-1,0\right)$ , so use $x=-1$ and $y=0$ to evaluate the trigonometric functions.

  

Now, using the above definition,

  $\begin{array}{c}\sin \theta =y\\ =0\\ \cos \theta =x\\ =-1\\ \csc \theta =\frac{r}{y}\\ =\cancel{\frac{1}{0}}\text{ undefined}\\ \sec \theta =\frac{r}{x}\\ =\frac{1}{-1}\\ =-1\\ \tan \theta =\frac{y}{x}\\ =\frac{0}{-1}\\ =0\\ \cot \theta =\frac{x}{y}\\ =\cancel{\frac{-1}{0}}\text{ undefined}\end{array}$