Trigonometrical Ratios And Identities Ques 27
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)
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Answer:
Correct Answer: 27.$(\frac{1}{64})$
Solution:
Formula:
- $\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14} \cdot \sin \frac{7 \pi}{14} \cdot \sin \frac{9 \pi}{14} \cdot \sin \frac{11 \pi}{14} \cdot \sin \frac{13 \pi}{14}$
$=\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14} \cdot \sin (\pi-\frac{5 \pi}{14})$ $\cdot \sin (\pi-\frac{3 \pi}{14}) \cdot \sin (\pi-\frac{\pi}{14})$
$=\sin ^{2} \frac{\pi}{14} \cdot \sin ^{2} \frac{3 \pi}{14} \cdot \sin ^{2} \frac{5 \pi}{14}=(\sin \frac{\pi}{14} \cdot \sin \frac{3 \pi}{14} \cdot \sin \frac{5 \pi}{14})^{2}$
$=(\cos (\frac{\pi}{2}-\frac{\pi}{14}) \cdot \cos (\frac{\pi}{2}-\frac{3 \pi}{14}) \cdot \cos (\frac{\pi}{2}-\frac{5 \pi}{14}))^2$
$=(\cos \frac{3 \pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{\pi}{7})^{2}$
$=(-\cos \frac{\pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7}{ })^{2}$
$=(-\frac{\sin 2^{3} \pi / 7}{2^{3} \cdot \sin \pi / 7})^2$
$=(-\frac{1}{8} \cdot \frac{\sin 8 \pi / 7}{\sin \pi / 7})^{2}$
$ \because \sin \frac{8 \pi}{7}=\sin (\pi+\frac{\pi}{7})=-\sin \frac{\pi}{7}$
$ =\frac{1} {64} $
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