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 - $cos2x=2cos^2x-1, sin2x=2sinxcosx$