Chapter 12.3, Problem 4WE

To determine

Give the values of six trigonometric functions of $\theta$ in standard form whose terminal side passes through the point $P(-8,-15)$

Answer to Problem 4WE

  $\sin \theta =\frac{-15}{17}$

  $\cos \theta =\frac{-8}{17}$

  $\tan \theta =\frac{15}{8}$

  $\csc \theta =\frac{17}{-15}$

  $\sec \theta =\frac{17}{-8}$

  $\cot \theta =\frac{8}{15}$

Explanation of Solution

Formula Used:

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

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

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

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

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

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

Calculation:

Since terminal side passes through the point $P(-8,-15)$

So,

x = length of the side adjacent to the angle $\theta$ = -8

y = length of the side opposite to the angle $\theta$ = -15

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

  $\sin \theta =\frac{y}{r}=\frac{-15}{17}$

  $\cos \theta =\frac{x}{r}=\frac{-8}{17}$

  $\tan \theta =\frac{y}{x}=\frac{-15}{-8}=\frac{15}{8}$

  $\csc \theta =\frac{r}{y}=\frac{17}{-15}$

  $\sec \theta =\frac{r}{x}=\frac{17}{-8}$

  $\cot \theta =\frac{x}{y}=\frac{8}{15}$