Closed forms of sums $f(a)+f(a+d)+\cdots+f(a+nd)$ with $f$ sine, cosine or tangent is/are there a closed form for
$\sin{(a)}+\sin{(a+d)}+\cdots+\sin{(a+n\,d)}$
$\cos{(a)}+\cos{(a+d)}+\cdots+\cos{(a+n\,d)}$
$\tan{(a)}+\tan{(a+d)}+\cdots+\tan{(a+n\,d)}$
$\sin{(a)}+\sin{(a^2)}+\cdots+\sin{(a^n)}$
$\sin{(\frac{1}{a})}+\sin{(\frac{1}{a+d})}+\cdots+\sin{(\frac{1}{a+n\,d})}$

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 $2\pi\mathbb{Z}$).
I do not think simple closed form formulas exist for the three other sums you are interested in.
