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cos(q)+ cos(3q) + cos(5q)+....+ cos(2n-1)q = sin(2nq)/sin(q)
To be archived: Complex variables problem
By Carolyn on Wednesday, September 26, 2001 - 12:26 am:

Please help!
Show that cos(q)+ cos(3q) + cos(5q)+....+ cos(2n-1)q = sin(2nq)/sin(q).

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

Multiply both sides by sin(q) 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 rq = (eirq+e-irq)/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(q)+sin(3q)+.....sin((2n-1)q)
then consider s+ic and apply euler's formula to each term. Sum the resulting gp and takethe real part of that. You have your answer.

abhaya