sin(2x+x)=sin2x.cosx+cos2x.sinx=2sinxcosx.cosx+(-2sin^2x)sinx=2sinxcos^2+sinx-2sin^3x=sinx(2cos^2x+1)-2sin^3x=sinx(2-2sin^2x+1)-2sin^3x=2sinx-2sin^3x+sinx-2sin^3x= 3sinx-4sin^3x

THE PROBLEM OR FORMULA IS RIGHT ,SO

sin3x=3sinx-4sin³x

 The derivation of the formula :- sin(3x)=3sinx - 4sin³x is absolutely correct . . . 

this is the derivation of sin3x.i thnk all the steps are correct.

sin 3x =3sin x -4sin³x

cos3x = 4 cos³x- 3 cosx