Trigonometrical Ratios and Identities 1 Question 24
24. The value of
$\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14} \cdot \sin \frac{7 \pi}{14} \cdot \sin \frac{9 \pi}{14} \cdot \sin \frac{11 \pi}{14} \cdot \sin \frac{13 \pi}{14}$
is equal to
(1991, 2M)
Analytical & Descriptive Questions
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Solution:
- $\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14} \cdot \sin \frac{7 \pi}{14} \cdot \sin \frac{9 \pi}{14} \cdot \sin \frac{11 \pi}{14} \cdot \sin \frac{13 \pi}{14}$ $=\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14} \cdot \sin \pi-\frac{5 \pi}{14}$ $\cdot \sin \pi-\frac{3 \pi}{14} \cdot \sin \pi-\frac{\pi}{14}$
$=\sin ^{2} \frac{\pi}{14} \cdot \sin ^{2} \frac{3 \pi}{14} \cdot \sin ^{2} \frac{5 \pi}{14}=\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14}^{2}$
$=\cos \frac{\pi}{2}-\frac{\pi}{14} \cdot \cos \frac{\pi}{2}-\frac{3 \pi}{14} \cdot \cos \frac{\pi}{2}-\frac{5 \pi}{14}$
$=\cos \frac{3 \pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{\pi}{7}^{2}$
$=-\cos \frac{\pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7}{ }^{2}$
$=-\frac{\sin 2^{3} \pi / 7}{2^{3} \cdot \sin \pi / 7}$
$=-\frac{1}{8} \cdot \frac{\sin 8 \pi / 7}{\sin \pi / 7}^{2} \quad \because \sin \frac{8 \pi}{7}=\sin \pi+\frac{\pi}{7}=-\sin \frac{\pi}{7}$
$$ =1 / 64 $$
