# Trig Power Series Question

• November 15th 2010, 12:13 AM
ihatemath
Trig Power Series Question
Any help on this would be greatly appreciated! Thank You!

http://i948.photobucket.com/albums/a...athproblem.jpg
• November 15th 2010, 12:32 AM
TheCoffeeMachine
Let $r = 2\theta$, then we have:

$\displaystyle\cos^2{\theta} = \frac{1}{2}\left(1+\cos{r}\right) = \frac{1}{2}\left(1+\sum_{n=0}^{\infty}\frac{(-1)^n(r)^{2n}}{(2n)!}\right)[/tex]

[tex]\displaystyle = \frac{1}{2}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n(r)^{2n}}{(2n)!} = \frac{1}{2}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n(2\theta)^{2n}}{(2n)!}$
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• November 15th 2010, 03:19 AM
ihatemath
Thank You very much! I didn't realize it was that simple! :D
• November 15th 2010, 06:34 AM
TheCoffeeMachine
Quote:

Originally Posted by ihatemath
Thank You very much! I didn't realize it was that simple! :D

It wasn't! The machine made it simple. (Wink) You're welcome!