$\cos (θ) + \cos (3 θ) + \cos (5 θ) + \dots + \cos ((2 n - 1) θ) = \sin (2 n θ) / \sin (θ)$

By Carolyn on Wednesday, September 26, 2001 - 12:26 am:

Please help!
Show that $\cos (θ) + \cos (3 θ) + \cos (5 θ) + \dots + \cos ((2 n - 1) θ) = \sin (2 n θ) / \sin (θ)$ .

By Jack Willis on Sunday, September 30, 2001 - 10:45 am:

Multiply both sides by $\sin (θ)$ and then use $\sin (A) \cos (B) = 1 / 2 (\sin (A + B) + \sin (A - B))$ and cancel terms.
Jack

By Kerwin Hui on Sunday, September 30, 2001 - 12:32 pm:

Another way is to use the equation

$\cos r θ = (e^{i r θ} + e^{- i r θ}) / 2$

to express $S$ as a geometric series, summing it and then using the similar expression for sin to give what you want.

Kerwin

By abhaya on Sunday, September 30, 2001 - 09:47 pm:

Same as the solution by kerwin but u can do like this
Take $c = \sin (θ) + \sin (3 θ) + \dots + \sin ((2 n - 1) θ)$

then consider $s + i c$ and apply Euler's formula to each term. Sum the resulting gp and takethe real part of that. You have your answer.

abhaya