Question

Find the exact values of the six trigonometric functions of $t$if $P=\left(-\frac{1}{3},\frac{2\sqrt{2}}{3}\right)$ is the point on the unit circle that corresponds to $t$.

Short answer

The value is

$$\begin{gathered} \tan t=-2\sqrt{2}, \\ \csc t=\frac{3\sqrt{2}}{4} \\ \sec t=-3 \\ \cot t=-\frac{\sqrt{2}}{4} \end{gathered}$$

Step-by-step solution

Given information

Given the point$(-\frac{1}{3},\frac{2\sqrt{3}}{3})$

Identify the point and use the formula

A point on circle is of the form $P(\sin \theta ,\cos \theta )$,so

$$\begin{gathered} \sin t=\frac{2\sqrt{2}}{3} \\ \cos t=-\frac{1}{3} \end{gathered}$$

$$\begin{gathered} \tan t=\frac{y}{x} \\ \tan t=\frac{\frac{2\sqrt{2}}{3}}{-\frac{1}{3}} \\ \tan t=-2\sqrt{2} \end{gathered}$$

Calculating other ratios

Calculating, we get

$$\begin{gathered} \csc t=\frac{1}{y} \\ csct=\frac{\frac{1}{2\sqrt{2}}}{3} \\ csct=\frac{3}{2\sqrt{2}}·\frac{\sqrt{2}}{\sqrt{2}} \\ csct=\frac{3\sqrt{2}}{4} \end{gathered}$$

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

Calculating other ratios

Calculating, we get

$$\begin{gathered} \cot t=\frac{x}{y} \\ \cot t=\frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}} \\ \cot t=-\frac{1}{2\sqrt{2}}·\frac{\sqrt{2}}{\sqrt{2}} \\ \cot t=-\frac{\sqrt{2}}{4} \end{gathered}$$