# Thread: Trig Power Series Question

1. ## Trig Power Series Question

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

2. 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)[/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)!}$
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3. Thank You very much! I didn't realize it was that simple!

4. Originally Posted by ihatemath
Thank You very much! I didn't realize it was that simple!
It wasn't! The machine made it simple. You're welcome!