Can you prove that ? cos2θ = 2cos^2θ-1 ......thanks

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Can you prove that ? cos2θ = 2cos^2θ-1 ......thanks

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Answer 1 · 2018-03-09T16:08:14

see below

Explanation:

We have, $cos 2 θ = cos (θ + θ)$ $cos2theta=cos(theta+theta)$
Or,applying $cos (a + b) = cos a cos b - sin a sin b$ $cos(a+b)=cosacosb-sinasinb$ ,we see,
$cos 2 θ = cos θ \times cos θ - sin θ \times sin θ$ $cos2theta=costheta xxcostheta-sinthetaxxsintheta$
= ${cos}^{2} θ - (1 - {cos}^{2} θ)$ $cos^2theta-(1-cos^2theta)$ {since ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$ }
= $2 {cos}^{2} θ - 1$ $2cos^2theta-1$ { $Hence p r o v e d$ $color(red)("Hence " proved)$ }

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Can you prove that ? cos2θ = 2cos^2θ-1 ......thanks

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