Using de Moivre's theorem demonstrate that "sin6x+sin2x(16(sinx)^4-16(sinx)^2+3)"

Consider, (cosX+isinX)⁶ using binomial expansion we find that this = "cos⁶X + 6icos⁵XsinX - 15cos⁴Xsin²X - 20icos³Xsin³X + 15cos²Xsin⁴X + 6icosXsin⁵X - sin⁶X". Next, by using de Moivre's theorem (r(cos(X)+isin(X)))ⁿ = rⁿ(cos(nX)+isin(nX)) we find that (cosX+isinX)⁶ is also = cos6X+isin6X. So in a brief summary the whole binomial expansion is = to cos6X+isin6X.Now, by looking at the imaginary parts of both sides we can see that "isin6X = 6icos⁵XsinX - 20icos³Xsin³X + 6icosXsin⁵X" I'm seeing some lovely factors that I'd love to pull out, so lets! (I've also removed the i as it was common on both sides)"sin6X = 2cosXsinX(3cos⁴X - 10cos²Xsin²X + 3sin⁴X)" by using our trig identities we know that 2cosXsinX = sin2X and that cos²X = 1 - sin²X so, "sin6X = sin2X(3(1 - sin²X)² - 10(1 - sin²X)sin²X + 3sin⁴X)" Expanding the brackets we end up with "sin6X = sin2X( 3 - 6sin²X + 3sin⁴X - 10sin²X + 10sin⁴X + 3sin⁴X)" which leaves us to our final line! that "sin6X = sin2X( 3 + 16sin⁴X - 16sin²X)" proved using de Moire's theorem