Compound And Multiple Angles

The key method is the telescoping cosine-product identity, which applies whenever the angles form a geometric progression that doubles, and it exploits the double-angle formula $\sin 2\theta =2\sin \theta \cos \theta$ to fold each factor into the next. The angles $20^{\circ }$, $40^{\circ }$, $80^{\circ }$ double successively, so the trick is to multiply and divide the whole product by $\sin 20^{\circ }$, forming $\frac{\sin 20^{\circ }\cos 20^{\circ }\cos 40^{\circ }\cos 80^{\circ }}{\sin 20^{\circ }}$. Applying the double-angle identity step by step, the leading pair $\sin 20^{\circ }\cos 20^{\circ }$ becomes $\frac{1}{2}\sin 40^{\circ }$; combining with the next cosine, $\frac{1}{2}\sin 40^{\circ }\cos 40^{\circ }=\frac{1}{4}\sin 80^{\circ }$; and once more, $\frac{1}{4}\sin 80^{\circ }\cos 80^{\circ }=\frac{1}{8}\sin 160^{\circ }$. Because $\sin 160^{\circ }=\sin (180^{\circ }-20^{\circ })=\sin 20^{\circ }$, the introduced denominator cancels, leaving $\frac{\frac{1}{8}\sin 20^{\circ }}{\sin 20^{\circ }}=\frac{1}{8}$. Option $\frac{1}{4}$ stops one doubling short of the full chain. Option $\frac{1}{16}$ applies the halving one extra time. Option $\frac{1}{2}$ ignores the chained telescoping structure entirely. This is the classic $\cos \theta \cos 2\theta \cos 4\theta$ product pattern frequently set in JEE Advanced. As a final numeric plausibility check, $0.94\times 0.77\times 0.17\approx 0.125$, which matches $\frac{1}{8}$ precisely.