Question

How do you prove `sin3x = 3sinx - 4sin^3x`?

Answer 1

Since the angle sum formula of sine is:

`sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta`,

and the double angle formula of cosine:

`cos(2alpha)=cos^2alpha-sin^2alpha=2cos^2alpha-1=1-2sin^2alpha`

then:

`sin(3x)=sin(2x+x)=sin(2x)cosx+cos(2x)sinx=`

`=(2sinxcosx)*cosx+(1-2sin^2x)sinx=`

`=2sinxcos^2x+sinx-2sin^3x=`

`=2sinx(1-sin^2x)+sinx-2sin^3x=`

`=2sinx-2sin^3x+sinx-2sin^3x=`

`=3sinx-4sin^3x`.

Answer 2

we can use DeMoivre's theorem

Explanation:

`(costheta+isintheta)^n=cosntheta+isinntheta--(1)`

since we want `sin^3x` we will expand `(cosx+isinx)^3`

`(cosx+isinx)^3=cos^3x+3cos^2(isinx)+3cosx(isinx)^2+(isinx)^3`

`(cosx+isinx)^3=cos^3x-3cosxsin^2x+i(3cos^2xsinx-sin^3x)`

from`(1)`

`cos3x+isin3x=cos^3x-3cosxsin^2x+i(3cos^2xsinx-sin^3x)`

equating imaginary parts

`sin3x=3cos^2xsinx-sin^3x`

but `cos^2x=1-sin^2x`

`sin3x=3(1-sin^2x)sinx-sin^3x`

`sin3x=3sinx-3sin^2xsinx-sin^3x`

`sin3x=3sinx-4sin^3x`

as required