Prove that-cos 7-cos 14-cos 28-cos 56-sin 68- 16 cos 83-

LHS=cos7 #176; #160;cos14 #176; #160;cos28 #176; #160;cos56 #176; On dividing and multiplying by 2sin7 #8728;, we get #160; #160; #160; ...

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Standard XII

Physics

Introduction

Prove that:co...

Question

Prove that:
$\cos 7 ° \cos 14 ° \cos 28 ° \cos 56 ° = \frac{\sin 68 °}{16 \cos 83 °}$

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Solution

$LHS = \cos 7 ° \cos 14 ° \cos 28 ° \cos 56 °$

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

$= \frac{1}{2 \sin 7 °} \times 2 \sin 7 ° \times \cos 7 ° \times \cos 14 ° \times \cos 28 ° \times \cos 56 ° = \frac{2 \sin 14 °}{2 \times 2 \sin 7 °} \times \cos 14 ° \times \cos 28 ° \times \cos 56 ° = \frac{2 \sin 28 °}{2 \times 4 \sin 7 °} \times \cos 28 ° \times \cos 56 °$

$= \frac{2 \sin 56 °}{2 \times 8 \sin 7 °} \times \cos 56 ° = \frac{\sin 112 °}{16 \sin 7 °} = \frac{\sin (180 ° - 68 °)}{16 \sin (90 ° - 83 °)} = \frac{\sin 68 °}{16 \cos 83 °} [∵ \sin (180 ° - θ) = sinθ & \sin (90 ° - θ) = cosθ] = RHS Hence proved .$

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Prove that: cos 7° cos 14° cos 28° cos 56°=sin 68°16 cos 83°

Prove that:
$\cos 7 ° \cos 14 ° \cos 28 ° \cos 56 ° = \frac{\sin 68 °}{16 \cos 83 °}$