Chapter 6, Problem 20RE

To determine

Find the exact value of the remaining trigonometric functions.

Answer to Problem 20RE

The value of the remaining trigonometric functions are

  $\begin{array}{l}\cos \theta =\frac{12}{13}\\ \tan \theta =-\frac{5}{12}\\ \cot \theta =-\frac{12}{5}\\ \csc \theta =-\frac{13}{5}\\ \sec \theta =\frac{13}{12}\end{array}$

Explanation of Solution

Given:

The given expressionis $\sin \theta =-\frac{5}{13}$ , $\frac{3\pi }{2}<\theta <2\pi$ .

Calculation:

  $\sin \theta =-\frac{5}{13}$

  $\frac{3\pi }{2}<\theta <2\pi$

  $\theta$ lies in the fourth quadrant.

Draw a unit circle diagram for trigonometric values.

  

Take clockwise as negative angle direction and counterclockwise as positive angle direction.

Use fundamental identities.

  $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ \Rightarrow {\cos }^2\theta =1-{\sin }^2\theta \\ \Rightarrow \cos \theta =+\sqrt{1-{\sin }^2\theta }\\ \Rightarrow \cos \theta =+\sqrt{1-{\left(-\frac{5}{13}\right)}^2}\\ \Rightarrow \cos \theta =+\sqrt{1-\frac{25}{169}}\\ \Rightarrow \cos \theta =+\sqrt{\frac{144}{169}}\\ \Rightarrow \cos \theta =+\frac{12}{13}\end{array}$

  $\begin{array}{l}\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{-\frac{5}{13}}{\frac{12}{13}}=-\frac{5}{13}\cdot \frac{13}{12}=-\frac{5}{12}\\ \cot \theta =\frac{1}{\tan \theta }=\frac{1}{-\frac{5}{12}}=-\frac{12}{5}\end{array}$

  $\begin{array}{l}\csc \theta =\frac{1}{\sin \theta }=\frac{1}{-\frac{5}{13}}=-\frac{13}{5}\\ \sec \theta =\frac{1}{\cos \theta }=\frac{1}{\frac{12}{13}}=\frac{13}{12}\end{array}$

Hence thevalue of the remaining trigonometric functions are

  $\begin{array}{l}\cos \theta =\frac{12}{13}\\ \tan \theta =-\frac{5}{12}\\ \cot \theta =-\frac{12}{5}\\ \csc \theta =-\frac{13}{5}\\ \sec \theta =\frac{13}{12}\end{array}$