Chapter 13.2, Problem 28WE

To determine

To Find: The exact values of the six trignometric functions of $s$.

Answer to Problem 28WE

The six trignometric functions of $s$ is ;

  $\sin s=\frac{12}{13}$$\cos s=\frac{-5}{13}$$\tan s=\frac{-12}{5}$$\cot s=-\frac{5}{12}$$\sec s=\frac{-13}{5}$$\cos ecs=\frac{13}{12}$

Explanation of Solution

Given information:

The given is $P\left(\frac{-5}{13},\frac{12}{13}\right)$ .

Calculation:

We have sin $s=y$ , and cos $s=x$ . Substitute the values of $x$ and $y$ in the formula,

  $\sin s=\frac{12}{13}$$\cos s=-\frac{5}{13}$

we know that tan $s=\frac{\sin s}{\cos s}$ if $\cos s\ne 0.$ substitute $\frac{12}{13}$ for $\sin s;$ and $-\frac{5}{13}$ for $\cos s$ in the formula, and simplify.

  $\begin{array}{c}\tan s=\frac{\frac{12}{13}}{-\frac{5}{13}}\\ =-\frac{12}{5}\end{array}$

We know that $\cot s=\frac{\cos s}{\sin s}$ if $\sin s\ne 0$ . Substitute $\frac{12}{13}$ for $\sin s,$ and $-\frac{5}{13}$ for $\cos s$ in the formula, and simplify.

  $\begin{array}{c}\cot s=\frac{-\frac{5}{13}}{\frac{12}{13}}\\ =-\frac{5}{12}\end{array}$

Similarly, find other trignometric functions.

  $\begin{array}{c}\sec s=\frac{1}{\cos s}\\ =\frac{1}{\frac{12}{13}}\\ =\frac{13}{12}\end{array}$$\begin{array}{c}\cos ecs=\frac{1}{\sin s}\\ =\frac{1}{\frac{-5}{13}}\\ =-\frac{13}{5}\end{array}$

Therefore, the six trignometric functions of $s$ is ;

  $\sin s=\frac{12}{13}$$\cos s=\frac{-5}{13}$$\tan s=\frac{-12}{5}$$\cot s=-\frac{5}{12}$$\sec s=\frac{-13}{5}$$\cos ecs=\frac{13}{12}$