Question

find the value of above question.​

Answer 1

According to question,

$\sf \: \sqrt{2 \sqrt{2+2cos4θ}} = 2cosθ$

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• squaring on both sides,

$\sf \implies \: 2 + \sqrt{2+2cos4θ} = 4 {cos}^{2}θ$

$\sf \implies \: \sqrt{2+2cos4θ} = 4 {cos}^{2}θ - 2$

$\sf \implies \: \sqrt{2+2cos4θ} = 2(2 {cos}^{2}θ - 1)$

$\sf \implies \: \sqrt{2+2cos4θ} = 2(cos2θ)$

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Again squaring,

$\sf \implies \: 2 + 2cos4θ = 4( {cos}^{2} 2θ)$

$\sf \implies \: 2cos4θ = 4( {cos}^{2}θ ) - 2$

$\sf \implies \: 2cos4θ = 2(cos4θ)$

$\therefore \sf LHS=RHS$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Hence \: proved$

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Hope it helps

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