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

1 Answer
Oct 25, 2015

See explanation.

Explanation:

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

[1]color(white)(XX)cos3x

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

Angle sum identity: cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta

[3]color(white)(XX)=cosxcos2x-sinxsin2x

Double angle identity: cos2alpha=2cos^2alpha-1

[4]color(white)(XX)=cosx(2cos^2x-1)-sinxsin2x

[5]color(white)(XX)=2cos^3x-cosx-sinxsin2x

Double angle identity: sin2alpha=2sinalphacosalpha

[6]color(white)(XX)=2cos^3x-cosx-sinx(2sinxcosx)

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

Pythagorean identity: sin^2alpha=1-cos^2alpha

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

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

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

Combine like terms.

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

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