Cos 7-cos 14-cos 28-cos 56-sin 68-16 cos 83-

Cos 7^∘ cos 14^∘ cos 28^∘ cos 56^∘ Divide and multiply by 2 sin 7^∘, we get 1/2 sin 7^∘ .2 sin 7^∘. cos 7^∘. cos 14^∘. cos 28^∘. cos 56^∘ =2 sin 14^∘/2.2 s ...

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Byju's Answer

Standard XII

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

cos 7∘cos 14∘...

Question

$c o s 7^{\circ} c o s 14^{\circ} c o s 28^{\circ} c o s 56^{\circ} = \frac{s i n 68^{\circ}}{16 c o s 83^{\circ}}$

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Solution

$c o s 7^{\circ} c o s 14^{\circ} c o s 28^{\circ} c o s 56^{\circ}$

Divide and multiply by 2 sin $7^{\circ}$ , we get

$\frac{1}{2 s i n 7^{\circ}} .2 s i n 7^{\circ} . c o s 7^{\circ} . c o s 14^{\circ} . c o s 28^{\circ} . c o s 56^{\circ} = \frac{2 s i n 14^{\circ}}{2.2 s i n 7^{\circ}} . c o s 14^{\circ} . c o s 28^{\circ} . c o s 56^{\circ} [∵ 2 s i n A c o s A = s i n 2 A] = \frac{2 s i n 28^{\circ}}{2.4 s i n 7^{\circ}} . c o s 28^{\circ} . c o s 56^{\circ} = \frac{2 s i n 56^{\circ}}{2.8 s i n 7^{\circ}} . c o s 56^{\circ} = \frac{s i n 112^{\circ}}{16 s i n 7^{\circ}} = \frac{s i n (180^{\circ} - 68)}{16 s i n (90^{\circ} - 83^{\circ})} = \frac{s i n 68^{\circ}}{16 c o s 83^{\circ}}$
= RHS

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Similar questions

Q. Prove that:

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

Prove that $cos A cos cos 2 A cos 4 A cos 8 A = \frac{sin 2^{4} A}{2^{4} sin A}$
Four can be used as a formula as given in and hence prove the following :
$cos 7^{\circ} cos 14^{\circ} cos 28^{\circ} cos 56^{\circ} = \frac{sin 68^{\circ}}{16 cos 83^{\circ}}$

cos68^o cos8^o + sin68^o sin8^o =?

Q. Evaluate

\frac{cos 56^{\circ}}{sin 34^{\circ}}

1 + c o s 56^{\circ} + c o s 58^{\circ} - c o s 66^{\circ} =

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cos 7∘ cos 14∘ cos 28∘ cos 56∘=sin 68∘16 cos 83∘

$c o s 7^{\circ} c o s 14^{\circ} c o s 28^{\circ} c o s 56^{\circ} = \frac{s i n 68^{\circ}}{16 c o s 83^{\circ}}$