Compound And Multiple Angles

Question

Using exact angle-sum reasoning, the value of the product $\cos 20^\circ \cdot \cos 40^\circ \cdot \cos 80^\circ$ simplifies to which of the following closed-form numbers?

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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 $\sin 2	heta = 2\sin	heta\cos	heta$ 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 $\cos	heta\cos 2	heta\cos 4	heta$ product pattern frequently set in JEE Advanced. As a final numeric plausibility check, $0.94 	imes 0.77 	imes 0.17 \approx 0.125$, which matches $\frac{1}{8}$ precisely.

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Compound And Multiple Angles

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Solution

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By repeatedly using the triple-angle identity, the exact value of $sin 3 θ$ when $\sin\theta = ...

Given that $tan^{- 1} x + tan^{- 1} y = \frac{π}{4}$ holds for positive reals with $x y < 1$ , the e...

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