find the exact value of each expression. (a) sin((31π/6)-(7π/3)) (b) sin(31π/6)-sin(7π/3)

Explanation:

Step1: Simplify the angles in part (a)

First, find a common - denominator for the angles inside the sine function in part (a). $\frac{31\pi}{6}-\frac{7\pi}{3}=\frac{31\pi - 14\pi}{6}=\frac{17\pi}{6}$. Then, $\frac{17\pi}{6}=2\pi+\frac{5\pi}{6}$. Since $\sin(x + 2k\pi)=\sin(x)$ for any real - number $x$ and integer $k$, $\sin(\frac{17\pi}{6})=\sin(\frac{5\pi}{6})$. And $\sin(\frac{5\pi}{6})=\frac{1}{2}$.

Step2: Evaluate the sine values in part (b)

For $\sin(\frac{31\pi}{6})$, $\frac{31\pi}{6}=5\pi+\frac{\pi}{6}$. Since $\sin(x + \pi)=-\sin(x)$, $\sin(5\pi+\frac{\pi}{6})=-\sin(\frac{\pi}{6})=-\frac{1}{2}$. For $\sin(\frac{7\pi}{3})$, $\frac{7\pi}{3}=2\pi+\frac{\pi}{3}$, and $\sin(2\pi+\frac{\pi}{3})=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$. Then $\sin(\frac{31\pi}{6})-\sin(\frac{7\pi}{3})=-\frac{1}{2}-\frac{\sqrt{3}}{2}$.

Answer:

(a) $\frac{1}{2}$ (b) $-\frac{1}{2}-\frac{\sqrt{3}}{2}$