Question

prove that cos 20degree cos 40degree cos 80degree =1/16
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Answer 1

Step-by-step explanation:

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Answer 2

Correct Question :-

$\bf \:Prove \: that \: cos20° cos40° cos80° = \dfrac{1}{8}$

Given :-

• cos20° cos40° cos80°

To prove :-

• cos20° cos40° cos80° = 1/8

Identity Used :

$\bf \:❥︎ \: 2cosAcosB = cos(A+ B) + cos(A - B)$

$\bf \:❥︎ \: cos(180° - θ) = - cosθ$

Solution :-

$\bf \:cos20° cos40° cos60°$

Multiply and divide by 2, we get

$\bf\implies \:\dfrac{1}{2} (2cos20° cos40° cos80°)$

$\bf\implies \:\dfrac{1}{2}(cos20° cos40°) cos80°$

$\bf\implies \:\dfrac{1}{2}(cos(20° + 40°) + cos(40° - 20°) )cos60°$

$\bf\implies \:\dfrac{1}{2}(cos60° + cos20°)cos80°$

$\bf\implies \:\dfrac{1}{2}( \dfrac{1}{2} + cos20°)cos80°$

$\bf\implies \:\dfrac{1}{2}( \dfrac{1}{2} cos80° + cos80°cos20°)$

Multiply and divide by 2, we get

$\bf\implies \:\dfrac{1}{4}(cos80° + 2cos80°cos20°)$

$\bf\implies \:\dfrac{1}{4}(cos80° + cos(80° + 20°) + cos(80° - 20°))$

$\bf\implies \:\dfrac{1}{4}(cos80° \: + cos 100°\: + cos60°)$

$\bf\implies \:\dfrac{1}{4}(cos80° + cos(180° - 80°) + \dfrac{1}{2} )$

$\bf\implies \:\dfrac{1}{4}(cos80° - cos80° + \dfrac{1}{2} )$

$\bf\implies \:\dfrac{1}{8}$

$\bf\implies \:cos20° cos40° cos80° = \dfrac{1}{8}$

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