Any help on this would be greatly appreciated! Thank You!

http://i948.photobucket.com/albums/a...athproblem.jpg

Printable View

- Nov 15th 2010, 12:13 AMihatemathTrig Power Series Question
Any help on this would be greatly appreciated! Thank You!

http://i948.photobucket.com/albums/a...athproblem.jpg - Nov 15th 2010, 12:32 AMTheCoffeeMachine
Let $\displaystyle r = 2\theta$, then we have:

$\displaystyle \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)[/Math]

[Math]\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)!} $. - Nov 15th 2010, 03:19 AMihatemath
Thank You very much! I didn't realize it was that simple! :D

- Nov 15th 2010, 06:34 AMTheCoffeeMachine