find the exact value of each expression. (a) sin(7π/12) - sin(π/12) = (b) sin(7π/24) cos(π/24) =

Explanation:

Step1: Use sum - to - product formula for (a)

The sum - to - product formula $\sin A-\sin B = 2\cos\frac{A + B}{2}\sin\frac{A - B}{2}$. Here $A=\frac{7\pi}{12}$ and $B = \frac{\pi}{12}$. Then $\frac{A + B}{2}=\frac{\frac{7\pi}{12}+\frac{\pi}{12}}{2}=\frac{\frac{8\pi}{12}}{2}=\frac{\pi}{3}$ and $\frac{A - B}{2}=\frac{\frac{7\pi}{12}-\frac{\pi}{12}}{2}=\frac{\frac{6\pi}{12}}{2}=\frac{\pi}{4}$. So $\sin\frac{7\pi}{12}-\sin\frac{\pi}{12}=2\cos\frac{\pi}{3}\sin\frac{\pi}{4}$. Since $\cos\frac{\pi}{3}=\frac{1}{2}$ and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, we have $2\times\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$.

Step2: Use double - angle formula for (b)

The double - angle formula $\sin 2\alpha=2\sin\alpha\cos\alpha$, so $\sin\alpha\cos\alpha=\frac{1}{2}\sin2\alpha$. Here $\alpha=\frac{\pi}{24}$, then $\sin\frac{7\pi}{24}\cos\frac{\pi}{24}=\frac{1}{2}\sin(\frac{7\pi}{24}+\frac{\pi}{24})=\frac{1}{2}\sin\frac{\pi}{3}$. Since $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, we get $\frac{1}{2}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$.

Answer:

(a) $\frac{\sqrt{2}}{2}$ (b) $\frac{\sqrt{3}}{4}$