Q. 14 - 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 $t$.

$\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$.

Step-by-Step Solution

Answer

The exact values of the six trigonometric functions of $t$ for $P=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ are:

$\begin{array}{c}\sin t=-\frac{\sqrt{3}}{2}\\ \cos t=\frac{1}{2}\\ \tan t=-\sqrt{3}\\ \csc t=-\frac{2}{3}\sqrt{3}\\ \sec t=2\\ \cot t=-\frac{1}{3}\sqrt{3}\end{array}$

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, $P\left(x,y\right)=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$.

Substitute these values in the six trigonometric functions and simplify. 

Find $\sin t$.

$\begin{array}{rcl}\sin t& =& y\\ \sin t& =& -\frac{\sqrt{3}}{2}\end{array}$

Find $\cos t$.

$\begin{array}{rcl}\cos t& =& x\\ \cos t& =& \frac{1}{2}\end{array}$

Find $\tan t$.

$\begin{array}{rcl}\tan t& =& \frac{y}{x}\\ \tan t& =& \frac{-{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}\\ \tan t& =& -\sqrt{3}\end{array}$

Evaluate.

Find $csct$.

$\begin{array}{rcl}csct& =& \frac{1}{y}\\ csct& =& \frac{1}{{\displaystyle \frac{-\sqrt{3}}{2}}}\\ csct& =& -\frac{2}{{\displaystyle \sqrt{3}}}\\ csct& =& -\frac{2}{{\displaystyle 3}}\sqrt{3}\end{array}$

Find $sect$.

$$\begin{gathered} sect=\frac{1}{x} \\ sect=\frac{1}{{\displaystyle \frac{1}{2}}} \\ sect=2 \end{gathered}$$

Find $cot t$.

$\begin{array}{rcl}cott& =& \frac{x}{y}\\ cott& =& \frac{{\displaystyle \frac{1}{2}}}{-{\displaystyle \frac{\sqrt{3}}{2}}}\\ cott& =& -\frac{1}{\sqrt{3}}\\ cott& =& -\frac{1}{3}\sqrt{3}\end{array}$