How do you solve the identity $cos 3 x = 4 {cos}^{3} x - 3 cos x$ $cos3x = 4cos^3x - 3cosx$ ?

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Answer 1 · 2015-10-25T04:31:14

See explanation.

So we will be proving that $cos 3 x = 4 {cos}^{3} x - 3 cos x$ $cos3x=4cos^3x-3cosx$

$[1] X X cos 3 x$ $[1]color(white)(XX)cos3x$

$[2] X X = cos (x + 2 x)$ $[2]color(white)(XX)=cos(x+2x)$

Angle sum identity: $cos (α + β) = cos α cos β - sin α sin β$ $cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta$

$[3] X X = cos x cos 2 x - sin x sin 2 x$ $[3]color(white)(XX)=cosxcos2x-sinxsin2x$

Double angle identity: $cos 2 α = 2 {cos}^{2} α - 1$ $cos2alpha=2cos^2alpha-1$

$[4] X X = cos x (2 {cos}^{2} x - 1) - sin x sin 2 x$ $[4]color(white)(XX)=cosx(2cos^2x-1)-sinxsin2x$

$[5] X X = 2 {cos}^{3} x - cos x - sin x sin 2 x$ $[5]color(white)(XX)=2cos^3x-cosx-sinxsin2x$

Double angle identity: $sin 2 α = 2 sin α cos α$ $sin2alpha=2sinalphacosalpha$

$[6] X X = 2 {cos}^{3} x - cos x - sin x (2 sin x cos x)$ $[6]color(white)(XX)=2cos^3x-cosx-sinx(2sinxcosx)$

$[7] X X = 2 {cos}^{3} x - cos x - {sin}^{2} x (2 cos x)$ $[7]color(white)(XX)=2cos^3x-cosx-sin^2x(2cosx)$

Pythagorean identity: ${sin}^{2} α = 1 - {cos}^{2} α$ $sin^2alpha=1-cos^2alpha$

$[8] X X = 2 {cos}^{3} x - cos x - (1 - {cos}^{2} x) (2 cos x)$ $[8]color(white)(XX)=2cos^3x-cosx-(1-cos^2x)(2cosx)$

$[9] X X = 2 {cos}^{3} x - cos x - (2 cos x - 2 {cos}^{3} x)$ $[9]color(white)(XX)=2cos^3x-cosx-(2cosx-2cos^3x)$

$[10] X X = 2 {cos}^{3} x - cos x - 2 cos x + 2 {cos}^{3} x$ $[10]color(white)(XX)=2cos^3x-cosx-2cosx+2cos^3x$

Combine like terms.

$[11] X X = 4 {cos}^{3} x - 3 cos x$ $[11]color(white)(XX)=4cos^3x-3cosx$

$color(blue)( :.cos3x=4cos^3x-3cosx)$

How do you solve the identity cos3x = 4cos^3x - 3cosxcos3x=4cos3x−3cosxcos3x = 4cos^3x - 3cosx?