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Prove that: ,20^{},,40^{},,60^{},,80^{} = {3}{16}

Question

Prove that:
$sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ} = \frac{3}{16}$

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Prove that:

sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ} = \frac{3}{16}

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Text solutionVerified

L H S : sin 2 0^{o} sin 4 0^{o} sin 6 0^{o} sin 8 0^{o}

= sin 6 0^{o} (sin 2 0^{o} sin 4 0^{o}) sin 8 0^{o}

= \frac{3}{2} \times \frac{1}{2} (2 sin 2 0^{o} sin 4 0^{o}) sin 8 0^{o}

= \frac{3}{4} [[cos (4 0^{o} - 2 0^{o}) - cos (4 0^{o} + 2 0^{o})] sin 8 0^{o}]

= \frac{3}{4} [cos (2 0^{o} - cos 6 0^{o}) sin 8 0^{o}]

= \frac{3}{4} [(cos 2 0^{o} - \frac{1}{2}) sin 8 0^{o}]

= \frac{3}{8} [2 cos 2 0^{o} sin 8 0^{o} - sin 8 0^{o}]

= \frac{3}{8} [sin 10 0^{o} + sin 6 0^{o} - sin 8 0^{o}]

= \frac{3}{8} [sin (18 0^{o} - 8 0^{o}) + \frac{3}{2} - sin 8 0^{o}]

= \frac{3}{8} \times \frac{3}{2} = \frac{3}{16}

= RHS

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Question Text	Prove that: $sin 2 0^{\circ} sin 4 0^{\circ} sin 6 0^{\circ} sin 8 0^{\circ} = \frac{3}{16}$
Answer Type	Text solution:1
Upvotes	150