Question 1034445
{{{cos(3x) + isin(3x) = (cosx+isinx)^3 = cos^3(x)+3cos^2(x)*isinx + 3cosx*(isinx)^2 + (isinx)^3}}},after applying de Moivre's and direct expansion.

= {{{cos^3(x)+3icos^2(x)sinx - 3cosx*sin^2(x) - isin^3(x)}}}.

Now equate the imaginary parts of the first complex quantity and the last complex quantity.

==> {{{sin(3x) = 3cos^2(x)sinx - sin^3(x) =  3(1-sin^2(x))sinx - sin^3(x) }}}

==>  {{{sin(3x) = 3sinx - 3sin^3(x) - sin^3(x) = 3sinx - 4sin^3(x)}}}

==> {{{sin(3x) = 3sinx - 4sin^3(x)}}}, or {{{highlight(4sin^3(x) = 3sinx - sin(3x))}}}.

As an added bonus, what would happen if you equated the real parts of the first complex quantity and the last complex quantity?  Try it out yourself.