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Prove the following:

cos 7° cos 14° cos 28° cos 56°= \(\frac {sin \,68°}{16 \,cos \,83°}\)

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L.H.S. = cos 7° cos 14° cos 28° cos 56°

\(\frac{1}{2\,sin\,7°}\)(2sin 7°cos 7°)cos 14°cos 28°cos 56°)

\(\frac{1}{2\,sin\,7°}\)(sin 14° cos 14° cos 28° cos 56°)

…[∵ 2sinθ cosθ = sin 2θ]
\(= \frac{1}{2\left(2 \sin 7^{\circ}\right)}\)(2sin 14° cos 14°) cos 28° cos 56°

\(= \frac{1}{4 \sin 7°}\)(sin 28° cos 28° cos 56°)

\(= \frac{1}{2(4 \sin 7°)}\)(2 sin 28° cos 28°) cos 56°

\(= \frac{1}{8 \sin 7°}\)(sin 56° cos 56°)

\(= \frac{1}{8 \sin 7°}\)(2 sin 56° cos 56°)

\(= \frac{1}{16 \sin 7°}\)(sin 112°)

\(\frac{180°-68°)}{16 sin(90°-83°)}\)

\(\frac {sin\,68°}{16 \,cos\, 83°}\)

R.H.S.

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