trigonometry - Generalization of the sum of angles formula for any number of angles

In another question one user helped me prove that the sum of three angles was a multiple of 360 degrees with formulas for sine and cosine sums of three angles. The sine formula was:
$\sin⁡(α+β+γ)=\sin ⁡α\cos⁡\beta\cos ⁡γ+\cos ⁡α\sin ⁡β\cos ⁡γ+\cos ⁡α\cos ⁡β\sin ⁡γ-\sin ⁡α\sin ⁡β\sin ⁡γ$

I infer that the pattern for five angles is as shown below? For brevity, I'm using a shorthand, e.g. $\sin⁡(α+β+γ):s(a_1+a_2+a_3 )$ and $\sin ⁡α\cos ⁡\beta\cos ⁡γ∶s_1 c_2 c_3$. So, is the following the proper pattern for summing five angles?
$$s(a_1+a_2+a_3+a_4+a_5)=s_1 c_2 c_3 c_4 c_5+c_1 s_2 c_3 c_4 c_5+c_1 c_2 s_3 c_4 c_5+c_1 c_2 c_3 s_4 c_5+c_1 c_2 c_3 c_4 s_5-s_1 s_2 s_3 s_4 s_5$$

If so, I can also infer the pattern for cosine and use the patterns for any number of angles.

Answer

$s(a_1+a_2+a_3) = s_1c_2c_3+c_1s_2c_3+c_1c_2s_3-s_1s_2s_3$

$c(a_1+a_2+a_3) =$ $$s(\frac \pi 2 -(a_1+a_2+a_3)) \\ =s((\frac \pi 2 -a_1)+(-a_2)+(-a_3)) \\= c_1c_2c_3 -s_1s_2c_3- s_1c_2s_3-c_1s_2s_3$$
And we also know ...
$s(a_4+a_5) = s_4c_5+ c_4s_5$

$c(a_4+a_5) = c_4c_5- s_4s_5$

So

$$s(a_1+a_2+a_3+a_4+a_5) \\ =( s_1c_2c_3+c_1s_2c_3+c_1c_2s_3-s_1s_2s_3) ( c_4c_5- s_4s_5) \\+ ( c_1c_2c_3 -s_1s_2c_3- s_1c_2s_3-c_1s_2s_3)(s_4c_5+ c_4s_5 )$$
There will be no cancellations so you will get 16 terms ( they will be the half of the $2^5=32$ possible arrangements of sines and cosines having an odd number of sines )

The coefficient in front of each term having $(2k+1)$ sines will be $(-1)^k$

So for 5 angles you would expect...

$\binom 51=5$ terms having 1 sine , coefficient = +1

$\binom 53=10$ terms having 3 sines , coefficient = -1

$\binom 55=1$ term having 5 sines , coefficient = +1