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I used TrigReduce on products of cosines in Mathematica.

It led to me to develop the formula $\prod_{i=1}^k\cos\left(\beta_i\right)=\sum_{j=1}^{2^{k-1}}\frac{\cos\left(P_j\right)}{2^{k-1}}, k \geq1$ where $P_j$ is a permutation of the signed sums of $\beta_i$'s, i.e. $\beta_1+\beta_2+\beta_3+\beta_4+\beta_6$ and $\beta_1-\beta_2+\beta_3-\beta_4+\beta_6$ are two of the 32 permutations with $k=6 $ ($\beta_1$ always has a positive coefficient).

How would I prove this? Is this formula out there in literature?

I tried proving it by using the chain rule and then reverse because integration is linear over sums, and derivatives make sine functions more "summy". Then was going to use a recursive formula to prove by induction. But that line of reasoning doesn't explain how the $\beta$'s get in the same cosinusoidal term. Also, nothing in my undergrad used this and I couldn't find anything on google. And Mathematica doesn't give the general formula when I use k instead of a fixed number. ( I also get something similar with sinusoidal products).

orangeskid
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    I'd write $\cos\beta_j=\frac12(e^{i\beta_j}+e^{-i\beta_j})$. – Angina Seng Oct 06 '17 at 16:34
  • I could easily write your comment into an answer, but I wouldn't have been able to figure it out for awhile without your help. If you write it out I will up it. –  Oct 06 '17 at 17:31
  • More or less all the trigonometric formulas follow from $\cos x=\frac12(e^{ix}+e^{-ix})$ – reuns Oct 06 '17 at 18:43

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Your formula is easily shown by indution. Suppose we have ($j_1=1$) \begin{eqnarray*} 2^{k-1} \prod_{i=1}^{k} \cos( \beta_i) = \sum_{j_2 \in \{1,-1\}} \cdots \sum_{j_k \in \{1,-1\}} \cos \left( \sum_{i=1}^{k} j_i \beta_i \right). \end{eqnarray*} Now multiply this by $2 \cos( \beta_{k+1}) $ and use $2 \cos(A) \cos(B) =\cos(A+B) +\cos(A-B)$. The summands on the RHS will become \begin{eqnarray*} 2 \cos \left( \sum_{i=1}^{k} j_i \beta_i \right) \cos ( \beta_{k+1} ) = \sum_{j_{k+1} \in \{1,-1\}} \cos \left( \sum_{i=1}^{k+1} j_i \beta_i \right). \end{eqnarray*}

orangeskid
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Donald Splutterwit
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