Chapter 6, Problem 21RE

To determine

Find the exact value of the remaining trigonometric functions.

Answer to Problem 21RE

The value of the remaining trigonometric functions are

  $\begin{array}{l}\cos \theta =-\frac{3}{\sqrt{10}}\\ \sin \theta =-\frac{1}{\sqrt{10}}\\ \cot \theta =3\\ \csc \theta =-\sqrt{10}\\ \sec \theta =-\frac{\sqrt{10}}{3}\end{array}$

Explanation of Solution

Given:

The given expressionis $\tan \theta =\frac{1}{3}$ , ${180}^{°}<\theta <{270}^{°}$ .

Calculation:

  $\tan \theta =\frac{1}{3}$

  ${180}^{°}<\theta <{270}^{°}$

  $\theta$ lies in the third quadrant.

Draw a unit circle diagram for trigonometric values.

  

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

Use fundamental identities.

  $\begin{array}{l}{\sec }^2\theta -{\tan }^2\theta =1\\ \Rightarrow \sec \theta =-\sqrt{1+{\tan }^2\theta }\\ \Rightarrow \sec \theta =-\sqrt{1+{\left(\frac{1}{3}\right)}^2}\\ \Rightarrow \sec \theta =-\sqrt{1+\frac{1}{9}}\\ \Rightarrow \sec \theta =-\sqrt{\frac{10}{9}}\\ \Rightarrow \sec \theta =-\frac{\sqrt{10}}{3}\end{array}$

  $\begin{array}{l}\cos \theta =\frac{1}{\sec \theta }=\frac{1}{-\frac{\sqrt{10}}{3}}=-\frac{3}{\sqrt{10}}\\ \cot \theta =\frac{1}{\tan \theta }=\frac{1}{\frac{1}{3}}=3\end{array}$

  $\begin{array}{l}\sin \theta =\tan \theta \cdot \cos \theta =\frac{1}{3}\cdot \left(-\frac{3}{\sqrt{10}}\right)=-\frac{1}{\sqrt{10}}\\ \csc \theta =\frac{1}{\sin \theta }=\frac{1}{-\frac{1}{\sqrt{10}}}=-\sqrt{10}\\ \sec \theta =\frac{1}{\cos \theta }=\frac{1}{-\frac{3}{\sqrt{10}}}=-\frac{\sqrt{10}}{3}\end{array}$

Hence thevalue of the remaining trigonometric functions are

  $\begin{array}{l}\cos \theta =-\frac{3}{\sqrt{10}}\\ \sin \theta =-\frac{1}{\sqrt{10}}\\ \cot \theta =3\\ \csc \theta =-\sqrt{10}\\ \sec \theta =-\frac{\sqrt{10}}{3}\end{array}$