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Find the sum of the series $\cos x + \cos 2x + \cdots + \cos (n-1)x.$ You must calculate the sum of this series only by multiplying through by $2\sin\left(\frac{x}{2}\right)$.

Now I've heard of finding the sum of a trig series by finding real and imaginary parts etc., but I have no idea how to do it this way.

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    Hint: $\cos(ax) \sin(x/2) = \dfrac{\sin((a+1/2)x) - \sin((a-1/2)x)}{2}$2012-12-30

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From the familiar identitiea $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$ we obtain $2\cos A\sin B=\sin(A+B)-\sin(A-B).$ Set $B=\sin(x/2)$ and $A=\cos(kx)$. We get $2\cos(kx)\sin(x/2)=\sin((k+1/2)x)-\sin((k-1/2)x).$ Add up, $k=1$ to $k=n-1$. There is a lot of cancellation (telescoping) on the right. The only surviving part is $\sin((n-1/2)x)-\sin(x/2)$.

Remark: The method doesn't quite work if $\sin(x/2)=0$, but then finding the sum of the cosines is easy. The closed-form formula is "almost" right even for these $x$, in the sense that it has the right limit.

A similar trick, using the addition law for cosine, gives a closed-form expression for $\sin x+\sin 2x+\cdots+\sin((n-1)x$.

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    Hi andre thank you :) would the limits for the sigma notation for this series be k=1 and n? could i go from there?2013-01-02