Chapter 4, Problem 80CR

To determine

To Find: To evaluate the given trigonometric functions of the given angle.

Answer to Problem 80CR

The required values are $0,1\text{and}0.$

Explanation of Solution

Given information: The given angle is $-3\pi$ .

Formula used: $\begin{array}{l}\mathrm{Sin}(2\pi +\theta )=\mathrm{Sin}\theta ,\mathrm{Cos}(2\pi +\theta )=\mathrm{Cos}\theta ,\mathrm{Tan}(2\pi +\theta )=\mathrm{Tan}\theta ,\\ \mathrm{Sin}(\pi +\theta )=-\mathrm{Sin}\theta ,\mathrm{Cos}(\pi +\theta )=-\mathrm{Cos}\theta ,\mathrm{Tan}(\pi +\theta )=\mathrm{Tan}\theta ,\\ \text{and}\mathrm{Sin}(-\theta )=-\mathrm{Sin}\theta ,\mathrm{Cos}(-\theta )=\mathrm{Cos}\theta ,\mathrm{Tan}(-\theta )=\mathrm{Tan}\theta \end{array}$ .

Calculation:

Since the given angle is $-3\pi$

Therefore, the angle lies in the III quadrant by moving clockwise.

Now, value of

  $\begin{array}{l}\sin (-3\pi )=-\sin (3\pi )\\ \sin (-3\pi )=-\sin (2\pi +\pi )\\ \sin (-3\pi )=-\sin (\pi )\\ \sin (-3\pi )=-\sin (\pi +0°)\\ \sin (-3\pi )=\sin (0°)\\ \therefore \sin (-3\pi )=0\end{array}$

Similarly, value of

  $\begin{array}{l}\cos (-3\pi )=\cos (3\pi )\\ \cos (-3\pi )=\cos (2\pi +\pi )\\ \cos (-3\pi )=\cos (\pi )\\ \cos (-3\pi )=\cos (\pi +0°)\\ \cos (-3\pi )=\cos (0°)\\ \therefore \cos (-3\pi )=1\end{array}$

and the value of

  $\begin{array}{l}\tan (-3\pi )=-\tan (3\pi )\\ \tan (-3\pi )=\tan (2\pi +\pi )\\ \tan (-3\pi )=\tan (\pi )\\ \tan (-3\pi )=\tan (\pi +0°)\\ \tan (-3\pi )=\tan (0°)\\ \therefore \tan (-3\pi )=0\end{array}$

Hence, the values are $0,1\text{and}0.$