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

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Answer 1 · 2016-06-06T17:18:39

Proof is given below.

$cos 3 θ = cos (2 θ + θ)$ $cos3theta=cos(2theta+theta)$

$= cos 2 θ cos θ - sin 2 θ sin θ$ $=cos2thetacostheta-sin2thetasintheta$

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

$= {cos}^{3} θ - {sin}^{2} cos θ - 2 {sin}^{2} θ cos θ$ $=cos^3theta-sin^2costheta-2sin^2thetacostheta$

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

$= cos θ ({cos}^{2} θ - 3 {sin}^{2} θ)$ $=costheta(cos^2theta-3sin^2theta)$

$= {cos}^{3} θ - 3 {sin}^{2} θ cos θ$ $=cos^3theta-3sin^2thetacostheta$

$= {cos}^{3} θ - 3 (1 - {cos}^{2} θ) cos θ$ $=cos^3theta-3(1-cos^2theta)costheta$

$= {cos}^{3} θ - 3 cos θ + 3 {cos}^{3} θ$ $=cos^3theta-3costheta+3cos^3theta$

$= 4 {cos}^{3} θ - 3 cos θ$ $=4cos^3theta-3costheta$

How do you prove cos3θ=4cos3θ−3cosθcos 3 theta = 4 cos^3 theta - 3 cos theta?