Help to sum these two series $\sin(x)-\sin(2x)+\sin(3x)...$

Question

I am reading Euler's paper entitled "Subsidium Calculi Sinuum" and he wrote down some "sums" for these trigonometric series:

\begin{align}S &= \sin(x)-\sin(2x)+\sin(3x)-\sin(4x)...\\\\ C &= \cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...\\\\ -1+C &= -1+\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...\\\\ -1+C+iS&=-1+[\cos(x)+\sin(x)]-[\cos(2x)+\sin(2x)+[\cos(3x)+\sin(3x)]-[\cos(4x)+\sin(4x)]...\end{align}

$$-1+e^{ix}-(e^{ix})^2+(e^{ix})^3-(e^{ix})^4+...$$

This is the geometric series, and I recognise that its "sum" is $-\dfrac{1}{1+e^{ix}}$

So, we have:

\begin{align} & -\frac{1}{1+e^{ix}}=-\frac{1+e^{-ix}}{(1+e^{ix})(1+e^{-ix})} \\[8pt] = {} & -\frac{1+\cos(x)-i\sin(x)}{(1+e^{ix})(1+e^{-ix})}=-\frac{1+\cos(x)-i\sin(x)}{2\cos(x)+2} \end{align}

Equating the real and imaginary part with $C+iS$, we have:

$$-1+C=-\frac{1+\cos(x)}{2+\cos(2x)}\implies C=\frac{1}{2}$$

$$S=-\frac{\sin(x)}{2\cos(x)+2}$$

However, for the sum of $S$, there shouldn't be a minus sign, according to Euler.

So what is wrong with my derivation?

Answer 1

Your numerator should be $1+e^{-ix}$, not $1+e^{ix}$.

Answer 2

Your error is here: $$ f(x)=\color{red}-\frac{1+\cos(x)\color{red}-i\sin(x)}{2\cos(x)+2}\implies S=\Im f(x)=\color{red}+\frac{\sin(x)}{2\cos(x)+2}. $$