Chapter 4, Problem 48RE

To determine

To find: the exact value of the remaining five trigonometric functions of the angle $\theta$ .

Answer to Problem 48RE

  $\begin{array}{l}\sin \theta =\frac{\text{2 }}{3},\cos \theta =\frac{-\sqrt{5}}{3},\tan \theta =\frac{-2}{-\sqrt{5}},\\ \cot \theta =\frac{-\sqrt{5}}{2},\sec \theta =\frac{-3}{-\sqrt{5}}\end{array}$

Explanation of Solution

Given information:

Consider $\csc \theta =\frac{3}{2},\text{ cos}\theta <0$

Calculation:

We have $\csc \theta =\frac{\text{hypotenuse}}{\text{opposite}}$

So, $\text{hypotenuse }=3,\text{apposite}=2$

Find the third side using Pythagorean theorem.

  $\begin{array}{l}adjacent=\sqrt{3^2-2^2}\\ =\pm \sqrt{5}\\ =-\sqrt{5}(\because \cos \theta <0)\end{array}$

Find the trigonometric functions

  $\begin{array}{l}\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\text{2 }}{3}\\ \cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{-\sqrt{5}}{3}\\ \tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{-2}{-\sqrt{5}}\\ \cot \theta =\frac{\text{adjacent}}{\text{opposite}}=\frac{-\sqrt{5}}{2}\\ \sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{-3}{-\sqrt{5}}\end{array}$

Hence, the exact values of five trigonometric functions are determined.