Summing $ \sum _{k=1}^{n} k\cos(k\theta) $ and $ \sum _{k=1}^{n} k\sin(k\theta) $ I'm trying to find 

$$\sum _{k=1}^{n} k\cos(k\theta)\qquad\text{and}\qquad\sum _{k=1}^{n} k\sin(k\theta)$$

I tried working with complex numbers, defining $z=\cos(\theta)+ i \sin(\theta)$ and using De Movire's, but so far nothing has come up.
HINT:

$$\sum_{1\le k\le n}\left(\cos k\theta+i\sin k\theta\right)=\sum_{1\le k\le n}e^{i k\theta}=e^{i\theta}\cdot\frac{e^{in\theta}-1}
{e^{i\theta}-1}$$

Differentiate  wrt $\theta$ and equate the real &amp; the imaginary parts 

$$\text{On differentiation, the left hand becomes }\sum_{1\le k\le n}k\left(-\sin k\theta+i\cos k\theta\right)$$



Alternatively, if $S=\sum_{1\le k\le n}k\cdot a^k=a+2\cdot a^2+\cdots+(n-1)\cdot a^{n-1}+n\cdot a^n$

$a\cdot S=a^2+2\cdot a^3+\cdots+(n-1)\cdot a^n+n\cdot a^{n+1}$

$$\text{On subtraction,}(a-1)S=n\cdot a^{n+1}-(a+a^2+\cdots+a^{n-1}+a^n)=n\cdot a^{n+1}-\frac{a(a^n-1)}{a-1}$$

$$\implies S=\sum_{1\le k\le n}k\cdot a^k=\frac{n\cdot a^{n+1}}{a-1}-\frac{a(a^n-1)}{(a-1)^2}$$

Putting $a=e^{i\theta},$ 

$$S=\sum_{1\le k\le n}k\cdot e^{ik\theta}=\frac{n\cdot e^{i(n+1)\theta}}{e^{i\theta}-1}-\frac{e^{i\theta}(e^{in\theta}-1)}{(e^{i\theta}-1)^2}$$ (The derivative method should bring us here,too)

$$S=\frac{ne^{i(n+1)\theta}}{e^{i\frac\theta2}(e^{i\frac\theta2}-e^{-i\frac\theta2})}-\frac{e^{i\theta}(e^{in\theta}-1)}{\{e^{i\frac\theta2}(e^{i\frac\theta2}-e^{-i\frac\theta2})\}^2}$$

As $e^{ix}-e^{-ix}=2i\sin x,$  $$S=\frac{ne^{i\frac{(2n+1)\theta}2}}{(e^{i\frac\theta2}-e^{-i\frac\theta2})}-\frac{(e^{in\theta}-1)}{(e^{i\frac\theta2}-e^{-i\frac\theta2})^2}$$

$$S=\frac{n\left(\cos\frac{(2n+1)\theta}2+i\sin\frac{(2n+1)\theta}2\right)}{2i\sin\frac\theta2}-\frac{(\cos n\theta+i\sin n\theta-1)}{(2i\sin\frac\theta2)^2}$$

$$=\frac{\cos n\theta-1}{4\sin^2\frac\theta2}+\frac{n\sin \frac{(2n+1)\theta}2}{2\sin\frac\theta2}+i\left(\frac{\sin n\theta}{4\sin^2\frac\theta2}-\frac{n(\cos\frac{(2n+1)\theta}2}{2\sin\frac\theta2}\right)$$

Again, $$S=\sum_{1\le k\le n}k\cdot e^{ik\theta}=\sum_{1\le k\le n}k(\cos k\theta+i\sin k\theta)=\sum_{1\le k\le n}k\cdot\cos k\theta+i\sum_{1\le k\le n}k\cdot\sin k\theta$$

Now, equate the real &amp; the imaginary parts
All identities are coming from  Wikipedia page for trigonometric identities. The main identity that we start with is:
$$
\sum_{k=1}^n \cos(kx)= \frac{ \sin((n+\frac{1}{2}) x)}{2\sin(x/2)} -1
$$
Get a derivative with respect to $x$, and we have
\begin{eqnarray*}
\sum_{k=1}^n k \sin(kx)&amp;=&amp;  \frac{  \cos(x/2) \sin((n+\frac{1}{2})x)- (2n+1) \cos((n+\frac{1}{2})x) \sin(x/2)}{4\sin^2(x/2)}\\
&amp;=&amp;  \frac{-2n \cos((n+\frac{1}{2})x) \sin(x/2) + \left(  \cos(x/2) \sin((n+\frac{1}{2})x)- \cos((n+\frac{1}{2})x) \sin(x/2)\right)}{4\sin^2(x/2)}\\
&amp;=&amp;  \frac{-2n \cos((n+\frac{1}{2})x) \sin(x/2) +   \sin(nx)}{4\sin^2(x/2)}\\
&amp;=&amp; \frac{ -n\sin( (n+1)x) + n\sin(nx) +    \sin(nx)}{4\sin^2(x/2)}\\
&amp;=&amp; \frac{ -n\sin( (n+1)x) + (n + 1)   \sin(nx)}{4\sin^2(x/2)}.
\end{eqnarray*}