Chapter 12.2, Problem 5WE

To determine

Find the values of the six trigonometric functions of an angle $\theta$ in standard position whose terminal side passes through point P.

Answer to Problem 5WE

  $\sin \theta =\frac{15}{17}$

  $\cos \theta =\frac{8}{17}$

  $\tan \theta =\frac{15}{8}$

  $\csc \theta =\frac{17}{15}$

  $\sec \theta =\frac{17}{8}$

  $\cot \theta =\frac{8}{15}$

Explanation of Solution

Given:

  $P(8,15)$

Formula Used:

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

Calculation:

In the triangle ,

x = length of side adjacent to $\theta$ = 8 units

y = length of side opposite to $\theta$= 15 units

r = length of the hypotenuse = $\sqrt{x^2+y^2}=\sqrt{64+225}=\sqrt{289}=17$

So,

  $\begin{array}{l}\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{y}{r}=\frac{15}{17}\\ \Rightarrow \sin \theta =\frac{15}{17}\end{array}$

  $\begin{array}{l}\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{x}{r}=\frac{8}{17}\\ \Rightarrow \cos \theta =\frac{8}{17}\end{array}$

  $\begin{array}{l}\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{15}{8}\\ \Rightarrow \tan \theta =\frac{15}{8}\end{array}$

  $\csc \theta =\frac{1}{\sin \theta }=\frac{17}{15}$

  $\sec \theta =\frac{1}{\cos \theta }=\frac{17}{8}$

  $\cot \theta =\frac{1}{\tan \theta }=\frac{8}{15}$