Question

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

Answer 1

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)`