# Deduce Q

• Oct 25th 2009, 10:46 PM
vuze88
Deduce Q
Given $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$

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

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

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.

thanks
• Oct 26th 2009, 02:24 PM
Opalg
Quote:

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

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

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

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.

$1+2\cos\theta+2\cos2\theta+...+2\cos n\theta= e^0 + (e^{i\theta}+ e^{-i\theta}) + \ldots + (e^{in\theta} + e^{-in\theta}) = \sum_{k=-n}^ne^{ik\theta}.$ This is a geometric series with common ratio $e^{i\theta}$ and sum $\frac{e^{-in\theta}(e^{i(2n+1)\theta}-1)}{e^{i\theta}-1}$. Divide top and bottom of that by $2ie^{i\theta/2}$ and you will have something that you can express in terms of sines.

For the second part (which should have an extra term +2n as shown), take the formula for the first part, replace $\theta$ by $2\theta$ and use the formula $\cos2\phi = 2\cos^2\phi-1$.