Chapter 6, Problem 18RE

To determine

Find the exact value of the remaining trigonometric functions.

Answer to Problem 18RE

The value of the remaining trigonometric functions are

  $\begin{array}{l}\csc \theta =\frac{5}{3}\\ \cos \theta =-\frac{4}{5}\\ \sin \theta =\frac{3}{5}\\ \tan \theta =-\frac{3}{4}\\ \cot \theta =\frac{5}{12}\end{array}$

Explanation of Solution

Given:

The given expressionis $\sec \theta =-\frac{5}{4}$ , $\tan \theta <0$ .

Calculation:

  $\sec \theta =-\frac{5}{4}$

  $\sec \theta <0$ and $\tan \theta <0$ , so $\theta$ lies in the second quadrant.

Draw a unit circle diagram for trigonometric values.

  

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

Use fundamental identities.

  $\begin{array}{l}\cos \theta =\frac{1}{\sec \theta }=\frac{1}{-\frac{5}{4}}=-\frac{4}{5}\\ \sin \theta =+\sqrt{1-{\cos }^2\theta }=+\sqrt{1-{\left(-\frac{4}{5}\right)}^2}=+\sqrt{\frac{9}{25}}=\frac{3}{5}\\ \cot \theta =\frac{1}{\tan \theta }=\frac{1}{\frac{12}{5}}=\frac{5}{12}\\ \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\frac{3}{5}}{-\frac{4}{5}}=\frac{3}{5}\cdot \left(-\frac{5}{4}\right)=-\frac{3}{4}\\ \csc \theta =\frac{1}{\sin \theta }=\frac{1}{\frac{3}{5}}=\frac{5}{3}\end{array}$

Hence thevalue of the remaining trigonometric functions are

  $\begin{array}{l}\csc \theta =\frac{5}{3}\\ \cos \theta =-\frac{4}{5}\\ \sin \theta =\frac{3}{5}\\ \tan \theta =-\frac{3}{4}\\ \cot \theta =\frac{5}{12}\end{array}$