Chapter 4.4, Problem 103E

(a)

To determine

To calculate:The two solutions of the equation $\sin \theta =\frac{1}{2}$ in both degrees and radians.

Answer to Problem 103E

The two solutions of the equation $\sin \theta =\frac{1}{2}$ are $\theta =30°,150°$ , and in radians $\theta =\frac{\pi }{6},\frac{5\pi }{6}$ .

Explanation of Solution

Given:

The trigonometric equation is $\sin \theta =\frac{1}{2}$ and $0°\le \theta <360°$ or $0\le \theta <2\pi$ in radians.

Calculation:

Since, the value of sin is $\frac{1}{2}$ and it gives the internal angles are $30°,60°$ and $90°$ . The opposite side is $1$ and the hypotenuse is $2$ . The value of base is $\sqrt{3}$ . Draw a triangle

  

So, the angle $\theta =30°$ and as per the trigonometric identity $\sin \left(180°-\theta \right)=\sin \theta$ , therefore another value of

  $\begin{array}{c}\theta =180°-30°\\ =150°\end{array}$ .

Convert $\theta =30°$ in radians

  $\begin{array}{c}\theta =30°\times \frac{\pi }{180°}\\ =\frac{\pi }{6}\end{array}$

For $\theta =150°$ in radians

  $\begin{array}{c}\theta =150°\times \frac{\pi }{180°}\\ =\frac{5\pi }{6}\end{array}$ .

Therefore the values of $\theta$ for $\sin \theta =\frac{1}{2}$ is $\theta =30°,150°$ in degrees and $\theta =\frac{\pi }{6},\frac{5\pi }{6}$ in radians.

(b)

To determine

To calculate:The two solutions of the equation $\sin \theta =-\frac{1}{2}$ in both degrees and radians.

Answer to Problem 103E

The two solutions of the equation $\sin \theta =-\frac{1}{2}$ are $\theta =210°,330°$ , and in radians $\theta =\frac{7\pi }{6},\frac{11\pi }{6}$ .

Explanation of Solution

Given:

The trigonometric equation is $\sin \theta =-\frac{1}{2}$ and $0°\le \theta <360°$ or $0\le \theta <2\pi$ in radians.

Calculation:

Since, the value of sin is $-\frac{1}{2}$ and by the trigonometric property $\sin \left(-x\right)=-\sin x$ and as $\sin 30°=\frac{1}{2}$ , therefore

  $\begin{array}{c}\sin \theta =-\sin 30°\\ =\sin \left(-30°\right)\end{array}$

So, the angle $\theta =-30°$ and as per the trigonometric identity $\sin \left(180°-\theta \right)=\sin \theta$ , therefore another value of

  $\begin{array}{c}\theta =180°-\left(-30\right)°\\ =180°+30°\\ =210°\end{array}$

Also the trigonometric identity $\sin \left(360°+\theta \right)=\sin \theta$ , therefore another value of $\begin{array}{c}\theta =360°+\left(-30\right)°\\ =360°-30°\\ =330°\end{array}$

Convert $\theta =210°$ in radians.

  $\begin{array}{c}\theta =210°\times \frac{\pi }{180°}\\ =\frac{7\pi }{6}\end{array}$

For $\theta =330°$ ,

  $\begin{array}{c}\theta =330°\times \frac{\pi }{180°}\\ =\frac{11\pi }{6}\end{array}$ .

Therefore the values of $\theta$ for $\sin \theta =-\frac{1}{2}$ is $\theta =210°,330°$ in degrees and $\theta =\frac{7\pi }{6},\frac{11\pi }{6}$ in radians.