Trigonometrical Ratios and Identities 1 Question 27
27. Without using tables, prove that
$\left(\sin 12^{\circ}\right)\left(\sin 48^{\circ}\right)\left(\sin 54^{\circ}\right)=\frac{1}{8}$.
(1982, 2M)
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Solution:
- $\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ}=\frac{1}{2}\left(2 \sin 12^{\circ} \sin 48^{\circ}\right) \sin 54^{\circ}$
$=\frac{1}{2}\left[\cos \left(36^{\circ}\right)-\cos \left(60^{\circ}\right)\right] \sin 54^{\circ}$
$=\frac{1}{2} \cos 36^{\circ}-\frac{1}{2} \sin 54^{\circ}$
$=\frac{1}{4}\left(2 \cos 36^{\circ} \sin 54^{\circ}-\sin 54^{\circ}\right)$
$=\frac{1}{4}\left(\sin 90^{\circ}+\sin 18^{\circ}-\sin 54^{\circ}\right)$
$=\frac{1}{4} 1+\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}$
$=\frac{1}{4} 1+\frac{\sqrt{5}-1-\sqrt{5}-1}{4}$
$$ =\frac{1}{4} 1-\frac{1}{2}=\frac{1}{8} $$
