Proof: ∑ [k=1...m] cos(2kΘ) = ( sin(mΘ) cos((m+1)Θ) ) / sin(Θ) .... m∈ℤ+, Θ ≠ nπ for ∀n∈ℤ?

I think the easiest way to treat this one is to first convert Cos(2kΘ) to the exponential form (1/2)(e^(-2ikΘ) + e^(2ikΘ)), and then find the infinite sum of a basic power series. Then simplify the resulting expression and convert back to trigonometric form. The intermediary form before conversion back to trigonometric form is:

(1/2) (1/(e^(2iΘ) - 1) (e^(-2iΘm)) (e^(2iΘm) - 1) (e^(2iΘ(m+1)) + 1)

From this, you can make out the trigonometric form.

Use Mathematical Induction

show that it is true for m = 1

Then assume m = W is true, with W ∈ℤ+

The show that m = W + 1 is also true

Then you will get to this part, but I seem to get stuck trying to prove it

sin(WΘ) cos((W+1)Θ) + sin(Θ) cos((2W+2)Θ) = sin((W+1)Θ) cos((W+2)Θ)

Damn.... my trigonometry sucks..... gonna star this so some math genius will prove this trigonometric identity.

Anyway, that's the idea how to solve it.

I'll be happier if there's a way to solve this without mathematical induction....