Prove that:


Prove that:

$\cos 7^{\circ} \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ}=\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}}$


LHS $=\cos 7^{\circ} \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ}$

On dividing and multiplying by $2 \sin 7^{\circ}$, we get

$=\frac{1}{2 \sin 7^{\circ}} \times 2 \sin 7^{\circ} \times \cos 7^{\circ} \times \cos 14^{\circ} \times \cos 28^{\circ} \times \cos 56^{\circ}$

$=\frac{2 \sin 14^{\circ}}{2 \times 2 \sin 7^{\circ}} \times \cos 14^{\circ} \times \cos 28^{\circ} \times \cos 56^{\circ}$

$=\frac{2 \sin 28^{\circ}}{2 \times 4 \sin 7^{\circ}} \times \cos 28^{\circ} \times \cos 56^{\circ}$

$=\frac{2 \sin 56^{\circ}}{2 \times 8 \sin 7^{\circ}} \times \cos 56^{\circ}$

$=\frac{\sin 112^{\circ}}{16 \sin 7^{\circ}}$

$=\frac{\sin \left(180^{\circ}-68^{\circ}\right)}{16 \sin \left(90^{\circ}-83^{\circ}\right)}$

$=\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}} \quad\left[\because \sin \left(180^{\circ}-\theta\right)=\sin \theta \& \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$


Hence proved. 

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