I was trying to evaluate the sums:
$$S_1=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$$ $$S_2=\sum_{n=1}^{\infty}\frac{\sin(nx+c)}{n}$$
I've found that $S_1=\tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right)$, however I can't seem to evaluate $S_2$. The problem is the $+c$ making it hard to simplify $\Im\left(e^{ixn}e^{ic}\right)$ with a similar technique.
To evaluate $S_1$, I said:
$$S_1=\sum_{n=1}^{\infty}\frac{\Im\left(e^{ix\cdot n}\right)}{n}=\Im\left(\sum_{n=1}^{\infty}\frac{\left(e^{ix}\right)^n}{n}\right)$$ Now, using the taylor series expansion of $\ln(1-a)$: $$-\ln(1-a)=\sum_{n=1}^{\infty}\frac{a^n}{n}$$ We have: $$S_1=\Im\left(-\ln\left(1-e^{ix}\right)\right)=\Im\left(-\ln\left(1-\cos(x)-i\sin(x)\right)\right)$$ Converting $1-\cos(x)-i\sin(x)$ into modulus-argument form: $$1-\cos(x)-i\sin(a)=\left(\sqrt{(1-\cos(x))^2+\sin^2(x)}\right)e^{i\tan^{-1}\left(\frac{-\sin(x)}{1-\cos(x)}\right)}$$ Hence: $$-\ln\left(1-\cos(x)-i\sin(x)\right)=-\ln\left(\sqrt{(1-\cos(x))^2+\sin^2(x)}\right)-i\tan^{-1}\left(\frac{-\sin(x)}{1-\cos(x)}\right)$$ So: $$S_1=\Im\left(-\ln\left(1-\cos(x)-i\sin(x)\right)\right)=-\tan^{-1}\left(\frac{-\sin(x)}{1-\cos(x)}\right)$$ Simplifying with trig identities yields: $$S_1=\tan^{-1}\left(\cot\left(\frac{x}{2}\right)\right)$$
