Cos 36-cos 42-cos 60-cos 78-1-16

(L.H.S. cos 36^∘ cos 42^∘ cos 60^∘ cos 78^∘ =1/2 cos 36^∘. cos 60^∘. (2 cos 42^∘. cos 78^∘) =1/2 (√(5)+1/4 ). 1/2 (cos 120^∘ + cos 36^∘) =(√(5)+1)/16( 1/2+√( ...

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Byju's Answer

Standard IX

Mathematics

Values of Trigonometric Ratios

cos 36∘cos 42...

Question

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

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Solution

$(L . H . S . c o s 36^{\circ} c o s 42^{\circ} c o s 60^{\circ} c o s 78^{\circ}$

$= \frac{1}{2} c o s 36^{\circ} . c o s 60^{\circ} . (2 c o s 42^{\circ} . c o s 78^{\circ})$

$= \frac{1}{2} (\frac{\sqrt{5} + 1}{4}) . \frac{1}{2} (c o s 120^{\circ} + c o s 36^{\circ})$

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

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

$= \frac{(\sqrt{5} + 1) (\sqrt{5} - 1)}{64}$

$= \frac{5 - 1}{64} = \frac{1}{16}$

=RHS

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Similar questions

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

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

Q. Prove that:

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

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

Q. Prove that:

\cos 78 ° \cos 42 ° \cos 36 ° = \frac{1}{8}

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cos 36∘ cos 42∘ cos 60∘ cos78∘=116

(L.H.S. cos 36∘ cos 42∘ cos 60∘ cos 78∘ =12 cos 36∘. cos 60∘. (2 cos 42∘. cos 78∘) =12 (√5+14). 12 (cos 120∘+cos 36∘) =(√5+1)16(−12+√5+14) =(√5+1)16(−2+√5+14) =(√5+1)(√5−1)64 =5−164=116 =RHS

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