Chapter 4.4, Problem 96E

(a)

To determine

Two solutions of $\sin \theta =\frac{\sqrt{3}}{2}$ in radians $0\le \theta <2\pi$ and in degrees $0^0\le \theta <{360}^0$ .

Answer to Problem 96E

  $\theta =\frac{\pi }{3}={60}^0\text{ and }\theta =\frac{2\pi }{3}={120}^0$

Explanation of Solution

Given information:

  $\sin \theta =\frac{\sqrt{3}}{2}$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant II sine functions are positive.

In quadrant III tangent functions are positive.

In quadrant IV cosine functions are positive.

Calculation:

Given that $\sin \theta =\frac{\sqrt{3}}{2}$ , Now,

  $\begin{array}{l}\sin \theta =\frac{\sqrt{3}}{2}\\ \theta =\frac{\pi }{3}={60}^0\text{ and }\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}={120}^0\end{array}$

Conclusion:

Thus,the solutions are $\theta =\frac{\pi }{3}={60}^0\text{ and }\theta =\frac{2\pi }{3}={120}^0$

(b)

To determine

Two solutions of $\sin \theta =-\frac{\sqrt{3}}{2}$ in radians $0\le \theta <2\pi$ and in degrees $0^0\le \theta <{360}^0$ .

Answer to Problem 96E

  $\theta =\frac{4\pi }{3}={240}^0\text{ and }\theta =\frac{5\pi }{3}={300}^0$

Explanation of Solution

Given information:

  $\csc \theta =\frac{2\sqrt{3}}{3}$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant II sine functions are positive.

In quadrant III tangent functions are positive.

In quadrant IV cosine functions are positive.

Calculation:

Given that $\sin \theta =-\frac{\sqrt{3}}{2}$ , Now,

  $\begin{array}{l}\sin \theta =-\frac{\sqrt{3}}{2}\\ \theta =\pi +\frac{\pi }{3}=\frac{4\pi }{3}={240}^0\text{ and }\theta =2\pi -\frac{\pi }{3}=\frac{5\pi }{3}={300}^0\end{array}$

Conclusion:

Thus,the solutions are $\theta =\frac{4\pi }{3}={240}^0\text{ and }\theta =\frac{5\pi }{3}={300}^0$