Rewrite the numerator as {[exp(2π) . exp(inπ)] - [exp(-2π) . exp(-inπ)]} Noting that (Eulers theorem) exp (iπ) = cos (π) + isin (π) = 1 exp(-iπ) = cos (π) - isin (π) = 1 and substituting The (-1) comes from putting n into Eulers formula, the result is alternately +1 and -1 Sorry the π (pi) symbols are almost indistinguishable from the n.
Damn spell checker again! I'll try not to scribble so quickly in future. you're right, it = -1 but that does work out.
Don't forget I rearranged your orignal expression by taking the -1 insde the bracket. It is part of the numerator. The last term in the third line should read exp minus (i n pi)