# closed form for some trigonometric sums

is/are there a closed form for

$\sin{(a)}+\sin{(a+d)}+\cdots+\sin{(a+n\,d)}$

$\cos{(a)}+\cos{(a+d)}+\cdots+\cos{(a+n\,d)}$

$\tan{(a)}+\tan{(a+d)}+\cdots+\tan{(a+n\,d)}$

$\sin{(a)}+\sin{(a^2)}+\cdots+\sin{(a^n)}$

$\sin{(\frac{1}{a})}+\sin{(\frac{1}{a+d})}+\cdots+\sin{(\frac{1}{a+n\,d})}$

$\ldots$

thanks

-
I was writing an answer, but Didier gave a much better one. But I will mention that Wolfram Alpha gives closed form for the first three. –  please delete me Mar 27 '11 at 9:40
@Elfvind Sorry about that. Re Wolfram Alpha, I would be at a loss to use in any way whatsoever the formula it provides for the tangent sum... –  Did Mar 27 '11 at 16:52
@Eivind: I misspelled your name. Sorry. I guess I linked somewhat unconsciously your name to Scandinavia and Scandinavia to elves... :-) –  Did Apr 2 '11 at 10:52
@Didier: Hehe, that's OK. =) –  please delete me Apr 4 '11 at 6:58

The two first sums $S$ and $C$ are the imaginary part and the real part of the complex sum $$\mathrm{e}^{\mathrm{i}a}+\mathrm{e}^{\mathrm{i}(a+d)}+\cdots+\mathrm{e}^{\mathrm{i}(a+nd)}.$$ Ths is nothing but a geometric sum with first term $\mathrm{e}^{\mathrm{i}a}$ and argument $z=\mathrm{e}^{\mathrm{i}d}$, thus $$C+\mathrm{i}S=\mathrm{e}^{\mathrm{i}a}\frac{z^{n+1}-1}{z-1}=\mathrm{e}^{\mathrm{i}(a+(nd/2))}\frac{\sin((n+1)d/2)}{\sin(d/2)},$$ for every $z\ne1$, that is, for every $d$ not in $2\pi\mathbb{Z}$. Using the shorthand $d=2b$ (hence $\sin(b)\ne0$), one gets finally $$C=\cos(a+nb)\frac{\sin((n+1)b)}{\sin(b)}, \quad S=\sin(a+nb)\frac{\sin((n+1)b)}{\sin(b)}.$$ If $d$ is in $2\pi\mathbb{Z}$, $C=(n+1)\cos(a)$ and $S=(n+1)\sin(a)$ (and these are the limits of the formulas valid when $d$ is not in $2\pi\mathbb{Z}$).