Finding the Value of Cos(2θ) from an Infinite Geometric Sequence

[physicsnoob93]

Alright, not really homework or coursework.

Sum(from k = 0 to infinity) (cos thetha)^2k = 5
So i subbed in the first value, 0 and took the one out.
I have to find the value of cos(2thetha)and then there was a pattern, but i didn't really know what to do.

Thanks

P.S. Not really familiar with Latex yet.

 

[dynamicsolo]

If you know the formula for the sum of an infinite series, you can work backwards. Substitute x = (cos theta)^2 to make your infinite sum

sum (k=0 to infinity) x^k = 5 .

What value of x works? That's your value for (cos theta)^2 . You can use a trig identity (or even a calculator) to find cos(2·theta)...

 

[HallsofIvy]

$\theta$ is a fixed number. As dynamicsolo said, first use the formula for the sum of an infinite geometric sequence
[itex]\sum_{k=0}^\infty x^k= \frac{1}{1- x}[/tex] to determine what x is. Then $\theta= cos^{-1}/sup](x)$.[/itex]