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Now use the two useful results given and proved by topsquark in this thread ( aha! I knew they'd be useful in the future )
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Now use the two useful results given and proved by topsquark in this thread ( aha! I knew they'd be useful in the future )
Thank you for the solution.
Since then I have seen another solution.
Let S = cosA + cos(A + B) + cos(A + 2B) + cos(A + 3B) + ... + cos(A + (n - 1)B).
2S sin(B/2) = sin(A + B/2) - sin(A - B/2) + sin(A + 3B/2) - sin(A + B/2) + sin(A + 5B/2) - sin(A + 3B/2) + ... + sin(A + (n - 1/2)B) - sin(A + (n - 3/2)B)
= -sin(A - B/2) + sin(A + (n - 1/2)B)
= 2cos(A + (n - 1)B/2)sin(nB/2).
sin(nB/2)
So S = ----------- . cos(A + (n - 1)B/2).
sin(B/2)