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If you look at the Taylor series expansion of e^x: e^x = 1 + x + x²/(2!) + x³/(3!) + x⁴/(4!)...

If you then make this e^ix, ...

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^...