Is there an identity to combine a sum of more than two sines; eg $\sin(a)+\sin(b)+\sin(c)+\sin(d)$? I get these trigonometric product to sum formulas like:
$$\sin(a)+\sin(b)=2\sin\frac12(a+b)\cos\frac12(a-b)$$
And that's useful, but I'm not too sure what to do if I need to turn a product into a sum if there's more than two variables.
What would I do with something like this?
$$\sin(a)+\sin(b)+\sin(c)+\sin(d)$$
The following formula works:
$$
\begin{align}
\sin(a)+sin(b)+sin(c)+sin(d) = 4*\sin\left(\frac{a+b+c+d}{4}\right)*\cos\left(\frac{a-b+c-d}{4}\right) \\
*\cos\left(\frac{a+b-c-d}{4}\right)*\cos\left(\frac{a-b-c+d}{4}\right)- \\
4*\cos\left(\frac{a+b+c+d}{4}\right)*\sin\left(\frac{a-b+c-d}{4}\right) \\
*\sin\left(\frac{a+b-c-d}{4}\right)*\sin\left(\frac{a-b-c+d}{4}\right) \\
\end{align}
$$
I wrote this Python script in Google colab to &quot;prove&quot; it:
    import numpy as np

    def C(x):
      return np.cos(180/np.pi*x)

    def S(x):
      return np.sin(180/np.pi*x)

    a = 37
    b = 6
    c = 88
    d = 7

    x = S(a) + S(b) + S(c) + S(d)
    y1 = 4*S((a+b+c+d)/4.0)*C((a-b+c-d)/4.0)*C((a+b-c-d)/4.0)*C((a-b-c+d)/4.0)
    y2 = 4*C((a+b+c+d)/4.0)*S((a-b+c-d)/4.0)*S((a+b-c-d)/4.0)*S((a-b-c+d)/4.0)
    print(x)
    print(y1-y2)

When the input is $(a,b,c,d) = (37,6,88,7)$, with angles measured in degrees, the outputs for $x$ and $(y1-y2)$ agree to $11$ decimal places at $-1.02707706592$.  I also tried $(a,b,c,d) = (10,27,18,68)$, and the two sides of the equation again agreed to $11$ decimal places. That isn't a mathematical proof, but the probability of my proposed identity holding true by accident for randomly selected angles is essentially nil. I tried making one of the angles obtuse, and it still worked.
Steps in the derivation:

Use angle sum formulas to derive an expression for $\sin\left(\frac{A+B+C+D}{4}\right)$ in terms of $\sin\left(\frac{A}{4}\right)$, $\sin\left(\frac{B}{4}\right)$, $\sin\left(\frac{C}{4}\right)$, $\sin\left(\frac{D}{4}\right)$, $\cos\left(\frac{A}{4}\right)$, $\cos\left(\frac{B}{4}\right)$, $\cos\left(\frac{C}{4}\right)$ and $\cos\left(\frac{D}{4}\right)$.

Use the first expression, and the fact that sine and cosine are odd and even functions, respectively, to derive expressions for $\sin\left(\frac{A-B+C-D}{4}\right)$, $\sin\left(\frac{A+B-C-D}{4}\right)$ and $\sin\left(\frac{A-B-C+D}{4}\right)$.

Add the four expressions together.

Make the following substitutions:
$$
A = a+b+c+d \\
B = a-b+c-d \\
C = a+b-c-d \\
D = a-b-c+d 
$$


...and you should get the formula.
The following formula works:
$$
\begin{align}
\sin(a)+sin(b)+sin(c)+sin(d) = 4*\sin\left(\frac{a+b+c+d}{4}\right)*\cos\left(\frac{a-b+c-d}{4}\right) \\
*\cos\left(\frac{a+b-c-d}{4}\right)*\cos\left(\frac{a-b-c+d}{4}\right)- \\
4*\cos\left(\frac{a+b+c+d}{4}\right)*\sin\left(\frac{a-b+c-d}{4}\right) \\
*\sin\left(\frac{a+b-c-d}{4}\right)*\sin\left(\frac{a-b-c+d}{4}\right) \\
\end{align}
$$
I wrote this Python script in Google colab to &quot;prove&quot; it:
    import numpy as np

    def C(x):
      return np.cos(180/np.pi*x)

    def S(x):
      return np.sin(180/np.pi*x)

    a = 37
    b = 6
    c = 88
    d = 7

    x = S(a) + S(b) + S(c) + S(d)
    y1 = 4*S((a+b+c+d)/4.0)*C((a-b+c-d)/4.0)*C((a+b-c-d)/4.0)*C((a-b-c+d)/4.0)
    y2 = 4*C((a+b+c+d)/4.0)*S((a-b+c-d)/4.0)*S((a+b-c-d)/4.0)*S((a-b-c+d)/4.0)
    print(x)
    print(y1-y2)

When the input is $(a,b,c,d) = (37,6,88,7)$, with angles measured in degrees, the outputs for $x$ and $(y1-y2)$ agree to $11$ decimal places at $-1.02707706592$.  I also tried $(a,b,c,d) = (10,27,18,68)$, and the two sides of the equation again agreed to $11$ decimal places. That isn't a mathematical proof, but the probability of my proposed identity holding true by accident for randomly selected angles is essentially nil. I tried making one of the angles obtuse, and it still worked.
Steps in the derivation:

Use angle sum formulas to derive an expression for $\sin\left(\frac{A+B+C+D}{4}\right)$ in terms of $\sin\left(\frac{A}{4}\right)$, $\sin\left(\frac{B}{4}\right)$, $\sin\left(\frac{C}{4}\right)$, $\sin\left(\frac{D}{4}\right)$, $\cos\left(\frac{A}{4}\right)$, $\cos\left(\frac{B}{4}\right)$, $\cos\left(\frac{C}{4}\right)$ and $\cos\left(\frac{D}{4}\right)$.

Use the first expression, and the fact that sine and cosine are odd and even functions, respectively, to derive expressions for $\sin\left(\frac{A-B+C-D}{4}\right)$, $\sin\left(\frac{A+B-C-D}{4}\right)$ and $\sin\left(\frac{A-B-C+D}{4}\right)$.

Add the four expressions together.

Make the following substitutions:
$$
A = a+b+c+d \\
B = a-b+c-d \\
C = a+b-c-d \\
D = a-b-c+d 
$$


...and you should get the formula.