Please help!
Show that cos(q)+cos(3q)+cos(5q)+¼+cos((2n-1)q) = sin(2nq)/sin(q).

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

cosrq = (e^{i rq}+e^{-i rq})/2

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

Kerwin

Same as the solution by kerwin but u can do like this
Take c=sin(q)+sin(3q)+¼+sin((2n-1)q)

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