How do you prove that ${cos}^{2} θ - {cos}^{2} 3 θ = sin 4 θ sin 2 θ$ $cos^2theta - cos^2 3theta = sin4thetasin2theta$ ?

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How do you prove that ${cos}^{2} θ - {cos}^{2} 3 θ = sin 4 θ sin 2 θ$ $cos^2theta - cos^2 3theta = sin4thetasin2theta$ ?

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Answer 1 · 2018-05-13T10:20:56

$L H S = {cos}^{2} θ - {cos}^{2} 3 θ$ $LHS=cos^2theta-cos^2 3theta$
$= \frac{1}{2} (1 + cos 2 θ) - \frac{1}{2} (1 + cos 6 θ)$ $=1/2(1+cos2theta)-1/2(1+cos6theta)$

$= \frac{1}{2} (cos 2 θ - cos 6 θ)$ $=1/2(cos2theta-cos6theta)$

$= \frac{1}{2} (2 sin (\frac{2 θ + 6 θ}{2}) sin (\frac{6 θ - 2 θ}{2}))$ $=1/2(2sin((2theta+6theta)/2)sin((6theta-2theta)/2))$

$= sin 4 θ sin 2 θ = R H S$ $=sin4thetasin2theta=RHS$

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How do you prove that ${cos}^{2} θ - {cos}^{2} 3 θ = sin 4 θ sin 2 θ$ $cos^2theta - cos^2 3theta = sin4thetasin2theta$ ?

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How do you prove that ${cos}^{2} θ - {cos}^{2} 3 θ = sin 4 θ sin 2 θ$ $cos^2theta - cos^2 3theta = sin4thetasin2theta$ ?