How do you prove that $cos θ - sin θ sin 2 θ = cos θ cos 2 θ$ $costheta-sinthetasin2theta=costhetacos2theta$ ?

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How do you prove that $cos θ - sin θ sin 2 θ = cos θ cos 2 θ$ $costheta-sinthetasin2theta=costhetacos2theta$ ?

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Answer 1 · 2018-05-26T10:16:10

$cos θ - sin θ sin 2 θ = cos θ - sin θ 2 sin θ cos θ =$ $costheta-sinthetasin2theta=costheta-sintheta2sinthetacostheta=$

$= cos θ - 2 {sin}^{2} θ cos θ = cos θ (1 - 2 {sin}^{2} θ)$ $=costheta-2sin^2thetacostheta=costheta(1-2sin^2theta)$

But we know that $cos 2 θ = {cos}^{2} θ - {sin}^{2} θ = 1 - {sin}^{2} θ - {sin}^{2} θ = 1 - 2 {sin}^{2} θ$ $cos2theta=cos^2theta-sin^2theta=1-sin^2theta-sin^2theta=1-2sin^2theta$

Then, we have $cos θ - sin θ sin 2 θ = cos θ cos 2 θ$ $costheta-sinthetasin2theta=costhetacos2theta$

QED

Answer 2 · 2018-05-26T10:22:49

$see explanation$ $"see explanation"$

Explanation:

$using the trigonometric identities$ $"using the "color(blue)"trigonometric identities"$

$∙ x sin 2 θ = 2 sin θ cos θ$ $•color(white)(x)sin2theta=2sinthetacostheta$

$∙ x cos 2 θ = 1 - 2 {sin}^{2} θ$ $•color(white)(x)cos2theta=1-2sin^2theta$

$consider the left side$ $"consider the left side"$

$cos θ - sin θ (2 sin θ cos θ)$ $costheta-sintheta(2sinthetacostheta)$

$= cos θ - 2 {sin}^{2} θ cos θ$ $=costheta-2sin^2thetacostheta$

$= cos θ (1 - 2 {sin}^{2} θ)$ $=costheta(1-2sin^2theta)$

$= cos θ cos 2 θ$ $=costhetacos2theta$

$= right side \Rightarrow verified$ $="right side "rArr"verified"$

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How do you prove that $cos θ - sin θ sin 2 θ = cos θ cos 2 θ$ $costheta-sinthetasin2theta=costhetacos2theta$ ?

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