If z = cisΘ, show that cos(nΘ) = 1/2 (z^n + (1/z^n)) I have done: z^n = (cisΘ)^n = cos(nΘ) + isin(nΘ) cos(nΘ) = z^n - isin(nΘ) Don't know where to go from here. Any help or advice would be greatly appreciated.
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Now do $\displaystyle z^{-n}$ and add it to what you've already got.
I've got it now =] Thank you very much
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