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

Question

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 1

In my opinion this formula is the best one: \begin{equation} \sin\left[\sum_{j=1}^N a_j\right] = \sum_{\sigma_{1,\dots,N-1}=\pm} \left[\prod_{j=1}^{N} \sin\left[a_j + \left(2 + \sigma_j-\sigma_{j-1} -\delta_{j1}(1-\sigma_{j-1}) - \delta_{jN}(1+\sigma_{j})\right)\frac{\pi}4\right]\right] \end{equation} The $\delta$'s are Kronecker deltas.

I hope it will be of use to someone; I used it to write the sine as a product of eigenvalues = a matrix. (This last fact is not relevant to the question, but a useful consequence of the above decomposition, i.e. any symbolic product can be written as a matrix with the individual factors of the product along the diagonal. The above produces a signed sum of these matrices, which gave the decomposition as given above some other application.)

EDIT:

Applied to the case N=5, the formula predicts \begin{align} \sin\left[\sum_1^5 a_j\right] = \sum_{\sigma_1,\sigma_2,\sigma_3,\sigma_4=\pm} \sin\left[a_1+(1+\sigma_1)\frac\pi4\right]\sin\left[a_2+(2+\sigma_2-\sigma_1)\frac\pi4\right]\sin\left[a_3+(2+\sigma_3-\sigma_2)\frac\pi4\right]\sin\left[a_4+(2+\sigma_4-\sigma_3)\frac\pi4\right]\sin\left[a_5+(1-\sigma_4)\frac\pi4\right] \end{align}

which can be verified by using the Mathematica code

pmlist = {-1, 1};
n = 5;
sigma\[LetterSpace]list\[LetterSpace]symb = 
  Reap[Do[Sow[
       ToExpression["\[Sigma]\[LetterSpace]" <> ToString[l]]], {l, 
       Range[n]}]][[2]][[1]];
sigma\[LetterSpace]list\[LetterSpace]symb\[LetterSpace]appended = \
{sigma\[LetterSpace]list\[LetterSpace]symb[[#]], pmlist} & /@ 
  Range[n - 1]
Sin[Sum[Subscript[a, i], {i, 1, n}]] - 
  Fold[Sum[#1, #2] &, 
   Product[Sin[
     Subscript[a, 
       j] + (\[Pi]/4)*(2 + 
         ToExpression["\[Sigma]\[LetterSpace]" <> ToString[j]] - 
         ToExpression["\[Sigma]\[LetterSpace]" <> ToString[j - 1]] - 
         KroneckerDelta[j, 
           1]*(1 - 
            ToExpression[
             "\[Sigma]\[LetterSpace]" <> ToString[j - 1]]) - 
         KroneckerDelta[j, 
           n] (1 + 
            ToExpression[
             "\[Sigma]\[LetterSpace]" <> ToString[j]]))], {j, 1, n}], 
   sigma\[LetterSpace]list\[LetterSpace]symb\[LetterSpace]appended] // \
TrigExpand

Answer 2

The correct formulas can be derived from the following knowledge. $$e^{i\alpha}=\cos\alpha + i\sin\alpha.$$

Hence $$\sin(a_1+a_2+a_3+a_4+a_5)=\renewcommand{\Im}{\operatorname{Im}}\Im e^{i(a_1+a_2+a_3+a_4+a_5)} = \Im e^{ia_1}e^{ia_2}e^{ia_3}e^{ia_4}e^{ia_5}$$ $$=\Im (c_1+is_1)(c_2+is_2)(c_3+is_3)(c_4+is_4)(c_5+is_5)$$ $$= \sum_{j=1}^5s_j\prod_{k\ne j} c_k - \sum_{\substack{\{j_1,j_2,j_3\}\subseteq[5]\\\{j_4,j_5\}=\{j_1,j_2,j_3\}^C}} s_{j_1}s_{j_2}s_{j_3} c_{j_4}c_{j_5} + s_1s_2s_3s_4s_5.$$

Note that the middle summation is the only piece missing from your formula above, and that the product of all sins has opposite sign from your formula. The reason we get this formula is that the only pieces of the product that contribute to the imaginary part are the pieces which have an odd number of sins in them. Thus we have three parts in our formula, since we can take 1 sin, 3 sins or 5 sins to get an imaginary piece.

In general, this leads one to derive the formula (correct me if my indices are messed up) $$\sin\left(\sum_{i=1}^n a_i\right) =\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\sum_{\substack{A\subseteq [n]\\|A|=2k+1}}(-1)^k\prod_{i\in A} s_i\prod_{j\in A^C}c_j.$$

Answer 3

$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

Answer 4

Alternatively, assuming that all angles $\,\alpha_j \in \left(0, 180^\circ\right)\,$ and none of them is a right angle, one can use that $\,\sum \alpha_j = k \cdot 360^\circ$ $\iff \sum \dfrac{\alpha_j}{2} = k \cdot 180^\circ$ $\iff \tan\left(\sum \dfrac{\alpha_j}{2}\right) = 0\,$.

For five angles $\,\alpha_j \mid j=1,2,3,4,5\,$, let $\,t_1 = \tan \dfrac{\alpha_1}{2}, t_2 = \tan \dfrac{\alpha_2}{2},\; \ldots \,, t_5 = \tan \dfrac{\alpha_1}{2}\,$, then using the $\tan$ of sum of angles formula:

$$ \tan\left(\frac{\alpha_1}{2}+\frac{\alpha_2}{2}+\ldots+\frac{\alpha_5}{2}\right) = \frac{e_1-e_3+e_5}{e_o-e_2+e_4} \quad\quad \style{font-family:inherit}{\text{where:}} \\[20px] \begin{align} \begin{cases} e_0 &= 1 \\ e_1 = \sum_{1 \le i \le 5} t_i &= t_1+t_2+t_3+t_4+t_5 \\ e_2 = \sum_{1 \le i \lt j \le 5} t_i &= t_1t_2+t_1t_3+t_1t_4+t_1t_5 \\ &\quad+t_2t_3+t_2t_4+t_2t_5 \\ &\quad+t_3t_4+t_3t_4 + \\ &\quad +t_4t_5 \\ e_3 = \sum_{1 \le i \lt j \lt k \le 5} t_it_jt_k &= t_1t_2t_3+t_1t_2t_4+t_1t_2t_5+t_1t_3t_4+t_1t_3t_5+t_1t_4t_5 \\ &\quad + t_2t_3t_4 + t_2t_3t_5+t_2t_4t_5 \\ &\quad + t_3t_4t_5 \\ e_4 = \sum_{1 \le i \lt j \lt k \lt l \le 5} t_it_jt_kt_l&= t_1t_2t_3t_4+t_1t_2t_3t_5+t_1t_2t_4t_5+t_1t_3t_4t_5+t_2t_3t_4t_5 \\ e_5 = \prod_{1 \le i \le 5} t_i &= t_1t_2t_3t_4t_5 \\ \end{cases} \end{align} $$

Answer 5

This questions asks for a simpler expression for $$ \sin\left(\sum_{i=1}^n \theta_i\right) $$ We will work on this using Euler's formula \begin{align} \sin\left(\sum_{i=1}^n \theta_i\right) =& \frac{1}{2i}\left(e^{i \sum \theta_i} - e^{-i\sum{\theta_i}}\right)\\ =& \frac{1}{2i}\left(\prod_i e^{i\theta_i} - \prod_ie^{-i\theta_i}\right)\\ =&\frac{1}{2i}\left(\prod_i \left(\cos(\theta_i) + i \sin (\theta_i)\right) - \prod_i \left(\cos(\theta_i) - i \sin(\theta_i)\right)\right)\\ \end{align} let $[n] = \{1, \ldots, n\} = \{m \in \mathbb{N}: 1 \le m \le n\}$. Then \begin{align*} =& \frac{1}{2i}\sum_{A\subset [n]}\left(\prod_{k\in A}(+i)\sin(\theta_k)\prod_{k \in [n]/A}\cos(\theta_k) - \prod_{k\in A}(-i)\sin(\theta_k)\prod_{k \in [n]/A}\cos(\theta_k)\right)\\ =& \frac{1}{2i}\sum_{A\subset[n]}\left(\left((+i)^{|A|} - (-i)^{|A|}\right)\prod_{k\in A}\sin(\theta_k)\prod_{k\in [n]/A}\cos(\theta_k)\right) \end{align*} $|A|$ is the cardinality of the set $A \subset [n]$. We have \begin{align} (+i)^{|A|} - (-i)^{|A|} = \begin{cases} 0 \text{ if }|A|\text{ is even}\\ 2i \text{ if }(|A|-1)/2 \text{ is even}\\ -2i \text{ if }(|A|-1)/2 \text{ is odd} \end{cases} \end{align} so we can write \begin{align} \sin\left(\sum_{i=1}^n \theta_i\right) = \sum_{\substack{A\subset[n] \\ |A|\text{ is odd}}} (-1)^{\frac{|A|-1}{2}} \prod_{k\in A}\sin(\theta_k)\prod_{k\in [n]/A}\cos(\theta_k) \end{align}

We can provide a formula for $\cos\left(\sum_{i=1}^n \theta_i\right)$ following the same logic. The differences are that the overall prefactor during the intermediate steps is $1/2$ instead of $1/2i$ and the two terms are added instead of subtracted so we instead have \begin{align} (+i)^{|A|} + (-i)^{|A|} = \begin{cases} 0 \text{ if }|A|\text{ is odd}\\ 2 \text{ if }|A|/2\text{ is even}\\ -2 \text{ if }|A|/2\text{ is odd} \end{cases} \end{align} so we get \begin{align} \cos\left(\sum_{i=1}^n \theta_i\right) = \sum_{\substack{A\subset[n] \\ |A|\text{ is even}}} (-1)^{\frac{|A|}{2}} \prod_{k\in A}\sin(\theta_k)\prod_{k\in [n]/A}\cos(\theta_k) \end{align} These expressions match those found on Wikipedia.

We can use this to calculate $$ \sin(\theta_1 + \theta_2 + \theta_3 + \theta_4 + \theta_5) $$ Here we have $[5] = \{1, 2, 3, 4, 5\}$. We must include the odd sized subsets of this set. There are 5 with size 1 that will have a + sign, 1 with size 5 that will have a + sign, and 10 with size 3 that will have a - sign. I don't have time to write them all out right now.