Cos 6-cos 42-cos 66-cos 78-1-16

L.H.S. = cos 6^∘ cos 42^∘ cos 66^∘ cos 78^∘ = 1/4 (2 cos 6^∘. cos 66^∘)(2 cos 42^∘. cos 78^∘) =1/4 (cos 72^∘ + cos 60^∘) (cos 120^∘ + cos 36^∘) =1/4(sin 1 ...

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Byju's Answer

Standard XII

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

cos 6∘cos 42∘...

Question

$c o s 6^{\circ} c o s 42^{\circ} c o s 66^{\circ} c o s 78^{\circ} = \frac{1}{16}$

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Solution

$L . H . S . = c o s 6^{\circ} c o s 42^{\circ} c o s 66^{\circ} c o s 78^{\circ} = \frac{1}{4} (2 c o s 6^{\circ} . c o s 66^{\circ}) (2 c o s 42^{\circ} . c o s 78^{\circ}) = \frac{1}{4} (c o s 72^{\circ} + c o s 60^{\circ}) (c o s 120^{\circ} + c o s 36^{\circ}) = \frac{1}{4} (s i n 18^{\circ} + \frac{1}{2}) (- \frac{2}{2} + \frac{\sqrt{5} + 1}{4}) = \frac{1}{4} (\frac{\sqrt{5} - 1}{4} + \frac{1}{2}) (\frac{\sqrt{5} + 1}{4} - \frac{1}{2})$

$= \frac{1}{4} (\frac{\sqrt{5} - 1 + 2}{4}) (\frac{\sqrt{5} + 1 - 2}{4}) = \frac{1}{64} (\sqrt{5} + 1) (\sqrt{5} - 1) = \frac{1}{64} ((\sqrt{5})^{2} - 1^{2}) = \frac{1}{64} (5 - 1) = \frac{1}{16}$
= RHS

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Similar questions

Question 1- sin 6 degree * sin 42 degree * sin 66 degree * sin 78 degree = ?

Question 2 - cos 6 degree* cos 42 degree * cos 66 degree * cos 78 degree =?

Q. Prove that:

\cos 6 ° \cos 42 ° \cos 66 ° \cos 78 ° = \frac{1}{16}

cos 60^{\circ} cos 36^{\circ} cos 42^{\circ} cos 78^{\circ} = \frac{1}{16}

$c o s 36^{\circ} c o s 42^{\circ} c o s 60^{\circ} c o s 78^{\circ} = \frac{1}{16}$

$c o s 78^{\circ} c o s 42^{\circ} c o s 36^{\circ} = \frac{1}{8}$

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cos 6∘ cos 42∘ cos 66∘ cos 78∘=116

$c o s 6^{\circ} c o s 42^{\circ} c o s 66^{\circ} c o s 78^{\circ} = \frac{1}{16}$