We have;
$$S=\sum_{k=0}^n \binom{n}{k} sin(kx)$$
I can think of two methods;
Method [1];
We can begin with $k=0,1,2,\cdots$ and consequently one can find a pattern and you will observe; $$S=2^nsin\left(\frac{nx}{2}\right)cos^n\left(\frac{x}{2}\right)$$
Method [2];
From Euler's Identity; $e^{ix}=isinx+cosx$, substitute $Sin(kx)= Im(e^{ikx})$,
$$S=\sum_{k=0}^n \binom{n}{k} Im(e^{ikx})$$
$$S=Im \left(\sum_{k=0}^n \binom{n}{k} (e^{ix})^k\right)$$
$$S=Im \left(1+e^{ix}\right)^n$$
$$S=Im \left(isinx+cosx+1\right)^n$$
$$S=Im \left(isinx+2cos^2\frac{x}{2}\right)^n$$
$$S=Im \left(2isin\frac{x}{2}cos\frac{x}{2}+2cos^2\frac{x}{2}\right)^n$$
$$S=2^nIm \left(isin\frac{x}{2}cos\frac{x}{2}+cos^2\frac{x}{2}\right)^n$$
$$S=2^n\,Im \left(e^{\frac{inx}{2}}cos^n\frac{x}{2}\right)$$
Using De Moivre theorem, and you can conclude here with
$$\sum_{k=0}^n \binom{n}{k} sin(kx)=2^nsin\left(\frac{nx}{2}\right)cos^n\left(\frac{x}{2}\right)$$
Extras - $cos2x=2cos^2x-1, sin2x=2sinxcosx$
The following standard identities are used in the answer:
$$\begin{align}
\frac{1}{1-x} &= \sum\limits_n x^n, \\
(x+y)^s &= \sum\limits_{i+j=s} \binom{i+j}{i} x^i y^j, \\
\frac{1}{(1-x)^{k+1}} &= \sum\limits_n \binom{n+k}{k} x^n.
\end{align}$$
To allow a better separation, let's say that we're given $n$ and $m$, and want to find
$$
\sum\limits_{i,j} \binom{i+j}{i} \binom{n-i}{j} \binom{m-j}{i}.
$$
This allows us to assume that $i$ and $j$ are fixed, and find the bivariate generating function
$$
\binom{i+j}{i}\sum\limits_{n=i}^\infty \binom{n-i}{j} x^n \sum\limits_{m=j}^\infty \binom{m-j}{i} y^m.
$$
Let's find the genfunc of individual sums via $\frac{1}{(1-x)^{k+1}} = \sum\limits_n \binom{n+k}{k} x^n$:
$$
\sum\limits_{n=i}^\infty \binom{n-i}{j} x^n = \frac{x^{i+j}}{(1-x)^{j+1}},
$$
thus the full bivariate genfunc of the sum is
$$
\sum\limits_{i=0}^\infty \sum\limits_{j=0}^\infty \binom{i+j}{i} \frac{x^{i+j}}{(1-x)^{j+1}}\frac{y^{i+j}}{(1-y)^{i+1}}.
$$
Using $s=i+j$, we rewrite the sum via $\sum\limits_{i+j=s} \binom{i+j}{i} x^i y^j = (x+y)^s$ and $\sum\limits_s x^s = \frac{1}{1-x}$ as
$$
\frac{1}{1-x}\frac{1}{1-y} \sum\limits_{s=0}^\infty \left(\frac{xy}{1-x}+\frac{xy}{1-y}\right)^s = \frac{1}{(1-xy)(1-x-y)}.
$$
What remains now is to find
$$\begin{align}
[x^n y^n]\frac{1}{(1-xy)(1-x-y)} &= [x^n y^n]\sum\limits_{k=0}^\infty x^k y^k \frac{1}{1-x-y} \\
&= \sum\limits_{k=0}^n [x^k y^k] \frac{1}{1-x-y} \\
&= \sum\limits_{k=0}^n [x^k y^k](x+y)^{2k} \\
&= \sum\limits_{k=0}^n \binom{2k}{k}. \square
\end{align}$$
P.S. It also shows that we can evaluate the sum for any $(n, m)$ as
$$\sum\limits_{i,j} \binom{i+j}{i} \binom{n-i}{j} \binom{m-j}{i} = \sum\limits_{k} \binom{n+m-2k}{n-k}.
$$
Best Answer
Hint: A well-known series (see here, e.g.) is:
$$\frac{1}{(1-x)^{11}} = \sum_{k=0}^\infty \binom{10+k}{k}x^k$$
Its derivative is close to what you want:
$$\frac{d}{dx} \frac{1}{(1-x)^{11}} = \sum_{k=1}^\infty k\binom{10+k}{k}x^{k-1}$$ At the end evaluate at $x=\frac12$ of course.