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
Related Further Mathematics A Level answers
All answers ▸Find the area of the surface generated when the curve with equation y=cosh(x) is rotated through 2 pi radians about the x axis, with 2<=x<=6
Prove that the sum of squares of the first n natural numbers is n/6(n+1)(2n+1)
FP3- Find the eigenvalues and the eigenvector for the negative eigenvalue, from this 2x2 matrix of columns (2,1) and (3,0)
You are given a polynomial f, where f(x)=x^4 - 14x^3 + 74 x^2 -184x + 208, you are told that f(5+i)=0. Express f as the product of two quadratic polynomials and state all roots of f.