Chapter 12.3, Problem 1WE

To determine

Give the values of six trigonometric functions of $\theta$ in standard form whose terminal side passes through the point $P(3,-4)$

Answer to Problem 1WE

  $\sin \theta =-\frac{4}{5}$

  $\cos \theta =\frac{3}{5}$

  $\tan \theta =-\frac{4}{3}$

  $\csc \theta =-\frac{5}{4}$

  $\sec \theta =\frac{5}{3}$

  $\cot \theta =-\frac{3}{4}$

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 $P(3,-4)$

So,

x = length of the side adjacent to the angle $\theta$ = 3

y = length of the side opposite to the angle $\theta$ = -4

r = length of the hypotenuse = $\sqrt{x^2+y^2}=\sqrt{16+9}=5$

  $\sin \theta =\frac{y}{r}=-\frac{4}{5}$

  $\cos \theta =\frac{x}{r}=\frac{3}{5}$

  $\tan \theta =\frac{y}{x}=-\frac{4}{3}$

  $\csc \theta =\frac{r}{y}=-\frac{5}{4}$

  $\sec \theta =\frac{r}{x}=\frac{5}{3}$

  $\cot \theta =\frac{x}{y}=-\frac{3}{4}$