Question

2sin($\frac{\pi}{22}$)sin($\frac{3\pi}{22}$)sin($\frac{5\pi}{22}$)sin($\frac{7\pi}{22}$)sin($\frac{9\pi}{22}$) is equal to

• A. $\frac{3}{16}$
• B. $\frac{1}{16}$
• C. $\frac{1}{32}$
• D. $\frac{9}{32}$

Answer 1

Answer: (B)

$\frac{1}{16}$

Answer 2

2sin($\frac{\pi}{22}$)sin($\frac{3\pi}{22}$)sin($\frac{5\pi}{22}$)sin($\frac{7\pi}{22}$)sin($\frac{9\pi}{22}$)
=2sin$\frac{(11\pi-10\pi)}{22}$sin$\frac{(11\pi-8\pi)}{22}$sin$\frac{(11\pi-6\pi)}{22}$sin$\frac{(11\pi-4\pi)}{22}$sin$\frac{(11\pi-2\pi)}{22}$
=2\frac{cos\pi}{11}$$\frac{2cos\pi}{11}$$\frac{cos3\pi}{11}$$\frac{cos4\pi}{11}$$\frac{cos5\pi}{11}
=$\frac{2sin\frac{32\pi}{11}}{2^5sin\frac{\pi}{11}}$=$\frac{1}{16}$