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 $t$.

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

Short answer

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

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

Step-by-step solution

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

Substitute these values in the six trigonometric functions and simplify.

Find $\sin t$.

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

Find $\cos t$.

$$\begin{gathered} \cos t=x \\ \cos t=-\frac{\sqrt{2}}{2} \end{gathered}$$

Find $\tan t$.

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

Evaluate.

Find $csct$.

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

Find $sect$.

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

Find $cott$.

$$\begin{gathered} cott=\frac{x}{y} \\ cott=\frac{-{\displaystyle \frac{\sqrt{2}}{2}}}{{\displaystyle \frac{\sqrt{2}}{2}}} \\ cott=-1 \end{gathered}$$