Chapter 12.3, Problem 19OE

To determine

Give the values of six trigonometric functions of $\theta$ . If any value is undefined , say so.

Answer to Problem 19OE

  $\sin \theta =\frac{-2}{\sqrt{13}}$

  $\cos \theta =\frac{-3}{\sqrt{13}}$$\tan \theta =\frac{2}{3}$

  $\csc \theta =\frac{\sqrt{13}}{-2}$

  $\sec \theta =\frac{\sqrt{13}}{-3}$

  $\cot \theta =\frac{3}{2}$

Explanation of Solution

Given:

  

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:

Given point $P(-3,-2)$

So,

x = length of the side adjacent to the angle θ= -3

y = length of the side opposite to the angle θ>= -2

r = length of the hypotenuse = $\sqrt{x^2+y^2}=\sqrt{9+4}=\sqrt{13}$

  $\sin \theta =\frac{y}{r}=\frac{-2}{\sqrt{13}}$

  $\tan \theta =\frac{y}{x}=\frac{-2}{-3}=\frac{2}{3}$

  $\csc \theta =\frac{r}{y}=\frac{\sqrt{13}}{-2}$

  $\sec \theta =\frac{r}{x}=\frac{\sqrt{13}}{-3}$

  $\cot \theta =\frac{x}{y}=\frac{3}{2}$