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De Moivre's Theorem states that if z = cos(q)+isin(q), then z n = (cos(q)+isin(q)) n = cos(nq)+isin(nq) But then z -n = cos(-nq)+isin(-nq). Now, cos(-p)=cos(p), as cosine is a symmetric (even) function, and ...

At GCSE level, proof questions are relatively rare and largely will all require a similar sort of approach. The difference with A Level is that the syllabus contains more than one method of proof. You will b...

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(...