Q.16 - 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 $t$.

$\left(-\frac{1}{5},\frac{2\sqrt{6}}{5}\right)$.

Step-by-Step Solution

Answer

The exact values of the six trigonometric functions of $t$ for $\left(-\frac{1}{5},\frac{2\sqrt{6}}{5}\right)$are:

$\begin{array}{c}\sin t=\frac{2\sqrt{6}}{5}\\ \cos t=-\frac{1}{5}\\ \tan t=-2\sqrt{6}\\ \csc t=\frac{5}{12}\sqrt{6}\\ \sec t=-5\\ \cot t=-\frac{1}{12}\sqrt{6}\end{array}$

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, $P\left(x,y\right)=\left(-\frac{1}{5},\frac{2\sqrt{6}}{5}\right)$.

Substitute these values in the six trigonometric functions and simplify.   

Find $\sin t$.

$$\begin{gathered} \sin t=y \\ \sin t=\frac{2\sqrt{6}}{5} \end{gathered}$$

Find $\cos t$.

$$\begin{gathered} \cos t=x \\ \cos t=-\frac{1}{5} \end{gathered}$$

Find $\tan t$.

$$\begin{gathered} \tan t=\frac{y}{x} \\ \tan t=\frac{{\displaystyle \frac{2\sqrt{6}}{5}}}{-{\displaystyle \frac{1}{5}}} \\ \tan t=-2\sqrt{6} \end{gathered}$$

Evaluate

Find$csct$.

$$\begin{gathered} csct=\frac{1}{y} \\ csct=\frac{1}{{\displaystyle \frac{2\sqrt{6}}{5}}} \\ csct=\frac{5}{{\displaystyle 2\sqrt{6}}} \\ csct=\frac{5}{{\displaystyle 12}}\sqrt{6} \end{gathered}$$

Find $sect$.

$$\begin{gathered} sect=\frac{1}{x} \\ sect=\frac{1}{-{\displaystyle \frac{1}{5}}} \\ sect=-5 \end{gathered}$$

Find $cott$.

$$\begin{gathered} cot t=\frac{x}{y} \\ cot t=\frac{-{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{2\sqrt{6}}{5}}} \\ cot t=-\frac{1}{2\sqrt{6}} \\ cot t=-\frac{1}{12}\sqrt{6} \end{gathered}$$