Generalizing trig-sum to product using complex exponentials Consider,
$$ \sin A \sin B$$
Using exponential definition of sine,
$$ \frac{ e^{iA} - e^{-iA} }{2i} \cdot \frac{ e^{iB} - e^{-iB} }{2i}$$
$$ =\frac{1}{-4}  ( e^{ i (A+B) } - e^{i (A-B)}  -e^{ -i(A-B) } + e^{ - i(A+B)})$$
$$ =\frac{-1}{4} ( 2\cos(A+B) - 2 \cos(A-B) )$$
or,
$$ \frac{1}{2} ( \cos(A-B) - \cos(A+B) )$$
Now, I want to generalize this trick for like
$$ \sin A \sin B \sin C \sin D....$$
 A: In more general terms we are being asked to express a product of sines and cosines as a sum of sines and cosines. So let us begin with the easiest case, that of a product of cosines and then see how to manipulate the answer in that case to deal with the product of sines case.
So let $\theta_1,\theta_2,\dots,\theta_n\in\mathbb{R}$ be the angles. We want to find a simplification for
$$
\prod_{k=1}^{n}\cos\theta_k=\frac{1}{2^n}\prod_{k=1}^{n}\left(\exp (i\theta_k)+\exp(-i\theta_k)\right).
$$
The difficulty is keeping track of the signs, so we introduce the notation $\epsilon=(\epsilon_1,\dots,\epsilon_n)$, where each $\epsilon_k\in\{-1,+1\}$. For convenience let us write $\theta:=(\theta_1,\dots,\theta_n)$. We then can sensibly use the compact notation $\epsilon\cdot\theta=\sum_{k=1}^{n}\epsilon_k\theta_k$. It is also convenient to use the compact notation $(-1)^\epsilon:=\epsilon_1\dots \epsilon_n$.
With this notation in place we see at once that
$$
\prod_{k=1}^{n}\left(\exp (i\theta_k)+\exp(-i\theta_k)\right)=
\sum_{\epsilon} \exp(i \epsilon\cdot\theta),
$$
where the sum is over the $2^n$ choices of $\epsilon$.
As the left hand side is real we need only take the real part of the right hand side, and so in fact have
$$
\prod_{k=1}^{n}\left(\exp (i\theta_k)+\exp(-i\theta_k)\right)=
\sum_{\epsilon} \cos(\epsilon\cdot\theta),
$$
or
$$
\prod_{k=1}^{n}\cos\theta_k=\frac{1}{2^n}\sum_{\epsilon} \cos(\epsilon\cdot\theta).
$$
To transform $\cos\theta_k$ to $\sin\theta_k$ we need only apply the operator $-\frac{\partial}{\partial\theta_k}$; and so we have
$$
\prod_{k=1}^{n}\sin\theta_k=\frac{(-1)^n}{2^n}\frac{\partial^n}{\partial\theta_1\dots\partial\theta_n}\sum_{\epsilon} \cos(\epsilon\cdot\theta).
$$
For the moment let us write $\cos^{(n)}(x)$ for the $n$-derivative of $\cos x$. Then we have
$$
\prod_{k=1}^{n}\sin\theta_k=\frac{(-1)^n}{2^n}\sum_{\epsilon}(-1)^{\epsilon} \cos^{(n)}(\epsilon\cdot\theta).
$$
Recall that $\cos'=-\sin$ and $\sin'=\cos$, which means that our answer depends on the congruence class of $n$ modulo $4$. So write $n=4m+r$ with $0\leqslant r<4$.
Finally we have that
$$
\textrm{ when $r=0$, } \prod_{k=1}^{n}\sin\theta_k=\frac{1}{2^n}\sum_{\epsilon}(-1)^{\epsilon} \cos(\epsilon\cdot\theta);
$$
$$
\textrm{ when $r=1$, } \prod_{k=-1}^{n}\sin\theta_k=\frac{1}{2^n}\sum_{\epsilon}(-1)^{\epsilon} \sin(\epsilon\cdot\theta);
$$
$$
\textrm{ when $r=2$,    } \prod_{k=-1}^{n}\sin\theta_k=\frac{-1}{2^n}\sum_{\epsilon}(-1)^{\epsilon} \cos(\epsilon\cdot\theta);
$$
$$
\textrm{ when $r=3$, } \prod_{k=-1}^{n}\sin\theta_k=\frac{-1}{2^n}\sum_{\epsilon}(-1)^{\epsilon} \sin(\epsilon\cdot\theta).
$$
The technique can be adapted to describe a product of $s$ cosines with $(n-s)$ sines; the problem is fixing on a compact enough notation to keep track of the sign changes and the $\cos/\sin$ swaps.
