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
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