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Let $S=S_1+S_2i$ $; S_1=\cos x+ \cos 2x + ...+ \cos nx $ $;S_2 \sin x + \sin 2x+...+ \sin nx $ and $z=\cos x + i\sin x$, then:

$$S=S_1+S_2i=z+z^2+...+z^n=z \frac{z^n-1}{z-1}=\frac{(\cos(n+1)x+\cos x)+i(\sin(n+1)x- \sin x)}{\cos x-1 + i\sin x -1}$$ Well, what I don't understand is this last line because I got something different: The only difference being that in the denominator being $\cos x+ i \sin x -1$ because $z-1?$ Then it goes on to say: $S_1= ReS$ and $S_2=Im S$ then (this I don't know how it is derived): $$S_1=\frac{\sin\frac{(n+1)x}{2}}{\sin \frac{x}{2}}\cos {(x+ \frac{nx}{2})} $$ $$S_2= \frac{\sin\frac{(n+1)x}{2} }{\sin \frac{x}{2}}\sin(x+\frac{nx}{2}) $$

Bozo Vulicevic
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  • See http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – lab bhattacharjee Sep 27 '15 at 15:56
  • If $S=S_1+S_2$, how do you later say $S=S_1+S_2i$? – Thomas Andrews Sep 27 '15 at 15:57
  • Also, $z-1\neq \cos x-1+i\sin x -1$. – Thomas Andrews Sep 27 '15 at 15:58
  • I think so too, that that is a typo in the book, am I write in my comment about that? – Bozo Vulicevic Sep 27 '15 at 16:00
  • Yes, looks like a typo on their part. – Thomas Andrews Sep 27 '15 at 16:01
  • I agree it is an error in the book. It might help if the question were clearer about which formulas are copied from a book and which are your own work. – David K Sep 27 '15 at 16:01
  • The next step is typical denominator simplification. If you have $a+bi$ in the denominator, multiply the numerator and denominator by $a-bi$. After you do that (and it will be messy,) you will need to use trig identities to get the result they have. – Thomas Andrews Sep 27 '15 at 16:03
  • Thanks a bunch, all this was from the book, except my correction of the denominator, I did stop doing my own work til' then because it was not coinciding with the book. Ill probably delete this now. :) – Bozo Vulicevic Sep 27 '15 at 16:05
  • Look at the answer http://math.stackexchange.com/a/1214626 -- I think it is an easier derivation of the real part of $S$ (though mostly quite similar to this way). – David K Sep 27 '15 at 16:08

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Write $z=\mathrm e^{\mathrm ix}$. Then: \begin{align*} S&=z\frac{z^n-1}{z-1}=\mathrm e^{\mathrm ix}\frac{\mathrm e^{\mathrm inx}-1}{\mathrm e^{\mathrm ix}-1}=\mathrm e^{\mathrm ix}\frac{\mathrm e^{\tfrac{\mathrm inx}2}\Bigl(\mathrm e^{\tfrac{\mathrm inx}2}-\mathrm e^{\tfrac{-\mathrm inx}2}\Bigr)}{\mathrm e^{\tfrac{\mathrm ix}2}\Bigl(\mathrm e^{\tfrac{\mathrm ix}2}-\mathrm e^{-\tfrac{\mathrm ix}2}\Bigr)}\\ &=\mathrm e^{\tfrac{\mathrm i(n+1)x}2}\frac{2\mathrm i\sin\dfrac{nx}2}{2\mathrm i\sin\dfrac x2}=\biggl(\cos\frac{(n+1)x}2+\mathrm i\sin\frac{(n+1)x}2\biggr)\frac{\sin\dfrac{nx}2}{\sin\dfrac x2}. \end{align*}

Bernard
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