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Question
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
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Solution
On dividing and multiplying by \[2\sin 7^\circ\] , we get
\[= \frac{1}{2\sin7^\circ } \times 2\sin7^\circ \times \cos7^\circ \times \cos14^\circ \times \cos28^\circ \times \cos56^\circ\]
\[ = \frac{2\sin14^\circ}{2 \times 2\sin7^\circ} \times \cos14^\circ \times \cos28^\circ \times \cos56^\circ \]
\[ = \frac{2\sin28^\circ}{2 \times 4\sin7^\circ} \times \cos28^\circ \times \cos56^\circ\]
\[ = \frac{\sin112^\circ}{16\sin7^\circ}\]
\[ = \frac{\sin\left( 180^\circ- 68^\circ\right)}{16\sin\left( 90^\circ - 83^\circ \right)}\]
\[ = \frac{\sin68^\circ}{16\cos83^\circ} \left[ \because \sin\left( 180^\circ - \theta \right) = sin\theta \sin\left( 90^\circ - \theta \right) = cos\theta \right]\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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