Question

C = 2cos(x) + 4cos(2x) + 8cos(3x) + ... + 2^n cos(nx)
S = 2sin(x) + 4sin(2x) + 8sin(3x) + ... + 2^n sin(nx)y

Show that:

Answer 1

\[C = \frac{ 2\cos(x) - 4 - 2^(n+1) \cos(n+1)x + 2^(n+2) cosnx }{ 5-4cosx }\]

Answer 2

the 5-4cosx I got from the previous question, i guess i'll insert that in. But apparantly C is a GP but i don't know how.

Answer 3

what is exactly the problem you wanna solve?

Answer 4

I need to show the thing that I posted as a comment, from the thing I posted as the question, and then eventually find a similar expression for S

Answer 5

And the thing you post as a comment has been calculated by you or makes part of the problem statement?

Answer 6

nope, that's a given statement, I just need to show it.

Answer 7

ok

Answer 8

so I found a similar sum online, do you think you could go through it and tell me if the method or whatever is right?? Them I could use that as reference to do this sum.
http://suite101.com/a/sum-of-trigonometric-series-using-de-moivres-theorem-cosxcos2x-a369345

Answer 9

let me have a look at it and will come back to you

Answer 10

thanks :)

Answer 11

It is a GP as you say because:\[C=\sum_{k=1}^{n}2^kcos(kx)=\sum_{k=1}^{n}2^k \frac{ e^{ikx}+e^{-ikx} }{ 2 }=\frac{ 1 }{ 2 }\sum_{k=1}^{n}\left[ 2e^{ix} \right]^k+\frac{ 1 }{ 2 }\sum_{k=1}^{n}\left[ 2e^{-ix} \right]^k\]which are two GP that have different common ratios. Let me see how to play with the ratios to find the final expression