Q. 20 - Precalculus Enhanced with Graphing Utilities

Question

In problems 13-20, $P\left(x,y\right)$ is the point on the unit circle that corresponds to a real number $t$. Find the exact values of the six trigonometric functions of $t$.

$\left(-\frac{\sqrt{5}}{3},-\frac{2}{3}\right)$.

Step-by-Step Solution

Answer

The exact values of the six trigonometric functions of $t$ for $\left(-\frac{\sqrt{5}}{3},-\frac{2}{3}\right)$ are:

$\begin{array}{c}\sin t=-\frac{2}{3}\\ \cos t=-\frac{\sqrt{5}}{3}\\ \tan t=\frac{2}{5}\sqrt{5}\\ \csc t=-\frac{3}{2}\\ \sec t=-\frac{3}{5}\sqrt{5}\\ \cot t=\frac{\sqrt{5}}{2}\end{array}$

Definitions.

For a point $P\left(x,y\right)$on a  unit circle that corresponds  to  $t$, the six trigonometric functions are defined as follows:  

$\begin{array}{c}\sin t=y\\ \cos t=x\\ \tan t=\frac{y}{x}\\ \csc t=\frac{1}{y}\\ \sec t=\frac{1}{x}\\ \cot t=\frac{x}{y}\end{array}$

Evaluate

Here, $P\left(x,y\right)=P\left(-\frac{\sqrt{5}}{3},-\frac{2}{3}\right)$.

Substitute these values in the six trigonometric functions and simplify.   

Find $\sin t$.

$$\begin{gathered} \sin t=y \\ \sin t=-\frac{2}{3} \end{gathered}$$

Find $\cos t$.

$$\begin{gathered} \cos t=x \\ \cos t=-\frac{\sqrt{5}}{3} \end{gathered}$$

Find $\tan t$.

$$\begin{gathered} \tan t=\frac{y}{x} \\ \tan t=\frac{-{\displaystyle \frac{2}{3}}}{-{\displaystyle \frac{\sqrt{5}}{3}}} \\ \tan t=\frac{2}{\sqrt{5}} \\ \tan t=\frac{2}{5}\sqrt{5} \end{gathered}$$

Evaluate.

Find $csc t$.

$$\begin{gathered} csc t=\frac{1}{y} \\ csc t=\frac{1}{-{\displaystyle \frac{2}{3}}} \\ csc t=-\frac{3}{2} \end{gathered}$$

Find $sec t$.

$$\begin{gathered} sect=\frac{1}{x} \\ sect=\frac{1}{-{\displaystyle \frac{\sqrt{5}}{3}}} \\ sec t=-\frac{3}{\sqrt{5}} \\ sec t=-\frac{3}{5}\sqrt{5} \end{gathered}$$

Find $cot t$.

$$\begin{gathered} cot t=\frac{x}{y} \\ cot t=\frac{-{\displaystyle \frac{\sqrt{5}}{3}}}{-{\displaystyle \frac{2}{3}}} \\ cot t=\frac{\sqrt{5}}{2} \end{gathered}$$