Chapter 12.3, Problem 51WE

To determine

Give the values of five other trigonometric functions of $\theta$ and find the quadrant.

Answer to Problem 51WE

  $\theta$ is in Quadrant II.

  $\cos \theta =\frac{-4}{5}$

  $\sin \theta =\frac{3}{5}$

  $\csc \theta =\frac{5}{3}$

  $\sec \theta =\frac{5}{-4}$

  $\cot \theta =\frac{-4}{3}$

Explanation of Solution

Given:

  $\tan \theta =-\frac{3}{4},\cos \theta <0$

Formula Used:

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

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

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

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

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

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

Function value	Quadrant of $\theta$ Is I	Quadrant of $\theta$ Is II	Quadrant of $\theta$ Is III	Quadrant of $\theta$ Is IV
$\begin{array}{l}\sin \theta \\ \csc \theta \end{array}$	+	+	-	-
$\begin{array}{l}\cos \theta \\ \sec \theta \end{array}$	+	-	-	+
$\begin{array}{l}\tan \theta \\ \cot \theta \end{array}$	+	-	+	-

Calculation:

  $\tan \theta =-\frac{3}{4},\cos \theta <0$

From the table ,

the angle $\theta$ is in Quadrant II, since tan and sec are negative in it.

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{y}{x}=\frac{3}{-4}$

So,

x = length of the side adjacent to the angle θ= -4, since Quadrant II

y = length of the side opposite to the angle θ= 3

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

  $\cos \theta =\frac{x}{r}=\frac{-4}{5}$

  $\sin \theta =\frac{y}{r}=\frac{3}{5}$

  $\csc \theta =\frac{r}{y}=\frac{5}{3}$

  $\sec \theta =\frac{r}{x}=\frac{5}{-4}$

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