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**vuze88** Given $\displaystyle \cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $\displaystyle \sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$

Deduce $\displaystyle 1+2\cos\theta+2\cos2\theta+...+2\cos n\theta=\frac{\sin(n+\frac{1}{2})\theta}{\sin\frac {1}{2}\theta}$

Use this result to show $\displaystyle 1+(2\cos\theta)^2+(2\cos2\theta)^2+...+(2\cos n\theta)^2=\frac{\sin(2n+1)\theta}{\sin\theta}\ {\color{red}{}+2n}$

I was thinking geometric series but cant seem to figure it out. It can easily be done using sums to products but i was wondering if there was a way which makes use of the given result.