Evaluate the following sum:  $n \choose 1$ $\sin (a) +$ $n \choose 2$  $\sin (2a) +...+$  $n \choose n$ $\sin (na)$

(A)  $2^n \cos^n \left(\frac{a}{2}\right)\sin \left(\frac{na}{2}\right)$

(B) $2^n \sin^n \left(\frac{a}{2}\right)\cos \left(\frac{na}{2}\right)$

(C) $2^n \sin^n \left(\frac{a}{2}\right)\sin \left(\frac{na}{2}\right)$

(D) $2^n \cos^n \left(\frac{a}{2}\right)\cos \left(\frac{na}{2}\right)$
To evaluate the sum 

\[
\sum_{k=1}^{n} \binom{n}{k} \sin(ka),
\]

we will use complex numbers and properties of binomial expansions. Let's define:

\[
s_n = \sum_{k=1}^{n} \binom{n}{k} \sin(ka),
\]

and also consider the sum involving cosine terms:

\[
c_n = \sum_{k=0}^{n} \binom{n}{k} \cos(ka).
\]

We can combine these sums into a single complex sum:

\[
c_n + i s_n = \sum_{k=0}^{n} \binom{n}{k} (\cos(ka) + i \sin(ka)).
\]

Using Euler's formula, we know that:

\[
\cos(ka) + i \sin(ka) = e^{ika}.
\]

Thus, the sum becomes:

\[
c_n + i s_n = \sum_{k=0}^{n} \binom{n}{k} e^{ika}.
\]

This is a binomial expansion of the form:

\[
\sum_{k=0}^{n} \binom{n}{k} (e^{ia})^k = (1 + e^{ia})^n.
\]

Next, we express \(e^{ia}\) using Euler's formula:

\[
e^{ia} = \cos(a) + i \sin(a).
\]

Therefore,

\[
1 + e^{ia} = 1 + \cos(a) + i \sin(a).
\]

We can simplify this expression using trigonometric identities. Notice that:

\[
1 + \cos(a) = 2 \cos^2\left(\frac{a}{2}\right),
\]

and

\[
\sin(a) = 2 \sin\left(\frac{a}{2}\right) \cos\left(\frac{a}{2}\right).
\]

Thus,

\[
1 + e^{ia} = 2 \cos^2\left(\frac{a}{2}\right) + i 2 \sin\left(\frac{a}{2}\right) \cos\left(\frac{a}{2}\right).
\]

Factoring out \(2 \cos\left(\frac{a}{2}\right)\), we get:

\[
1 + e^{ia} = 2 \cos\left(\frac{a}{2}\right) \left( \cos\left(\frac{a}{2}\right) + i \sin\left(\frac{a}{2}\right) \right).
\]

Recognizing that \(\cos\left(\frac{a}{2}\right) + i \sin\left(\frac{a}{2}\right) = e^{i \frac{a}{2}}\), we have:

\[
1 + e^{ia} = 2 \cos\left(\frac{a}{2}\right) e^{i \frac{a}{2}}.
\]

Raising this to the power \(n\), we get:

\[
(1 + e^{ia})^n = \left( 2 \cos\left(\frac{a}{2}\right) e^{i \frac{a}{2}} \right)^n = 2^n \cos^n\left(\frac{a}{2}\right) e^{i \frac{na}{2}}.
\]

Separating the real and imaginary parts, we have:

\[
c_n + i s_n = 2^n \cos^n\left(\frac{a}{2}\right) \left( \cos\left(\frac{na}{2}\right) + i \sin\left(\frac{na}{2}\right) \right).
\]

Thus, the imaginary part, which corresponds to \(s_n\), is:

\[
s_n = 2^n \cos^n\left(\frac{a}{2}\right) \sin\left(\frac{na}{2}\right).
\]

The final answer is \(\boxed{2^n \cos^n \left(\frac{a}{2}\right)\sin \left(\frac{na}{2}\right)}\).