Chapter 4, Problem 4CT

To determine

To Find:the exact value of 6 trigonometric function of the angle $\theta$ shown in figure

Answer to Problem 4CT

The value of $6$ trigonometric functions are

  $\begin{array}{l}\sin {120}^0=\frac{\sqrt{3}}{2}\\ \cos {120}^0=-\frac{1}{2}\\ \tan {120}^0=-\sqrt{3}\\ \sec {120}^0=-2\\ \cos ec{120}^0=\frac{2}{\sqrt{3}}\\ \cot {120}^0=-\frac{1}{\sqrt{3}}\end{array}$

Explanation of Solution

Given information:

The graph provided is shown below:-

Here x-axis and y-axis are on real number line and $\theta$ is an angle equivalent to ${120}^0$

Formula used:

The formula for trigonometric function are as follows

  $\begin{array}{l}\sin (\pi -\theta )=\sin \theta \\ \cos (\pi -\theta )=-\cos \theta \\ \tan \theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{1}{\tan \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ \cos ec\theta =\frac{1}{\sin \theta }\end{array}$

And the use of trigonometric table

  $\begin{array}{l}\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\\ \cos \frac{\pi }{3}=\frac{1}{2}\end{array}$

Calculation:

Consider the statement “ given trigonometric function”

Recall that $\theta$ is ${120}^0$ which is equal to $\frac{2\pi }{3}$

  $\begin{array}{l}\sin \frac{2\pi }{3}=\sin (\pi -\frac{\pi }{3})=\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}\\ \end{array}$

recall the formula of $\cos (\pi -\theta )$ is $-\cos \theta$

  $\cos \frac{2\pi }{3}=\cos (\pi -\frac{\pi }{3})=-\cos \frac{\pi }{3}=-\frac{1}{2}$

recall the formula of $\tan (\pi -\theta )$ is $\frac{\sin (\pi -\theta )}{\cos (\pi -\theta )}$

therefore,

  $\tan (\frac{2\pi }{3})=\tan (\pi -\frac{\pi }{3})=\frac{\sin (\pi -\frac{\pi }{3})}{\cos (\pi -\frac{\pi }{3})}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3}$

Recall the formula of $\sec \theta$ is $\frac{1}{\cos \theta }$

Therefore

  $\sec \frac{2\pi }{3}=\frac{1}{\cos \frac{2\pi }{3}}=-2$

Recall the formula of $\cos ec\theta$ is $\frac{1}{\sin \theta }$

Therefore , $\cos ec\frac{2\pi }{3}=\frac{1}{\sin \frac{2\pi }{3}}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$

Recall the formula of $\cot \theta$ is $\frac{1}{\tan \theta }$

  $\cot \frac{2\pi }{3}=\frac{1}{\tan \frac{2\pi }{3}}=-\frac{1}{\sqrt{3}}$

So,six trigonometric functions are

  $\begin{array}{l}\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}\\ \cos \frac{2\pi }{3}=-\frac{1}{2}\\ \tan \frac{2\pi }{3}=-\sqrt{3}\\ \cos ec\frac{2\pi }{3}=\frac{2}{\sqrt{3}}\\ \sec \frac{2\pi }{3}=-2\\ \cot \frac{2\pi }{3}=\frac{-1}{\sqrt{3}}\\ \end{array}$