Trigonometrical Ratios and Identities 1 Question 23
23. If $k=\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}$, then the numerical value of $k$ is
$(1993,2 M)$
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Solution:
- Using the relation,
$$ \sin \theta \sin \frac{\pi}{3}-\theta \sin \frac{\pi}{3}+\theta=\frac{\sin 3 \theta}{4} $$
Taking $\theta=\frac{\pi}{18}$, we get
$$ \sin \frac{\pi}{18} \cdot \sin \frac{5 \pi}{18} \cdot \sin \frac{7 \pi}{18}=\frac{\sin \frac{\pi}{6}}{4}=\frac{1}{8} $$
Alternative Method
Given, $\quad k=\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}$
$=\cos 80^{\circ} \cdot \cos 40^{\circ} \cdot \cos 20^{\circ}$
$=\cos A \cdot \cos 2 A \cdot \cos 2^{2} A=\frac{\sin 2^{3} A}{2^{3} \sin A}$
where, $A=20^{\circ}$
$$ =\frac{\sin 160^{\circ}}{8 \sin 20^{\circ}}=\frac{\sin \left(180^{\circ}-20^{\circ}\right)}{8 \sin 20^{\circ}}=\frac{\sin 20^{\circ}}{8 \sin 20^{\circ}}=\frac{1}{8} $$
