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If you look at the Taylor series expansion of e x: e x = 1 + x + x 2 /(2!) + x 3 /(3!) + x 4 /(4!)... If you then make this e ix , you get 1+ix - x 2 /(2!) - ix 3 /(3!) + x 4 /(4!)... If we split this into r...

The easiest way of approaching this question is to use De Moivre's formula: e^(inx) = cos(nx) + isin(nx) from which it is simple to show that cos(nx) = (e^(inx) + e^(-inx)) / 2 and sin(nx) = (e^(inx))- e^(-i...

By quoting De Moivre's theorem, (r(cos(x) + isin(x))) n = r n (cos(nx) + isin(nx)), we can realise that cos(4x) is a result of the real parts of (cos(x) + isin(x)) 4 Using binomial expansion: (where c = cos(...

We first write each side of the equation using the maclaurin series for each function. e ix = 1 + ix + (ix) 2 /2! + (ix) 3 /3! + (ix) 4 /4! + ...... e ix = 1 + ix - x 2 /2! - ix 3 /3! + x 4 /4! + ..... cos(x...

Two matrices are commutative is matrix A multiplied by matrix B equals matrix B multiplied by matrix A.