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I was wondering if there could be a general formula for this. Somehow, I managed to get (after some very small experimentation, so I'm not sure of what I'm going to say) that $$\sin(A_1+A_2+\cdots+A_n)=\cos A_1 \cos A_1 \cdots \cos A_n \left(\sum_{i=1}^n \tan A_i - \sum_{1\leq i<j<k\leq n} \left(\tan A_i \tan A_j \tan A_k\right)\right)$$

And I cannot prove it. Can anyone help me prove or disprove it?

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    $\begingroup$ Not sure, but induction on $n$ might be the way to go? $\endgroup$ – Shuri2060 Jul 18 '17 at 16:05
  • $\begingroup$ @Shuri2060 Maybe. I tried using induction but then I was encountering $\cos(A_1+A_2+\cdots+A_n)$ whose formula I'm not sure of, again. $\endgroup$ – Mathejunior Jul 18 '17 at 16:09
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It's not true for $n=5$: $$(\text{left side}) - (\text{right side}) = \sin(A_1)\sin(A_2)\sin(A_3)\sin(A_4)\sin(A_5) $$

I suspect that on the right side, for each $k$ with $2k+1 \le n$ you need $$(-1)^k \sum_{1 \le i_1 < \ldots < i_{2k+1}} \tan(A_{i_1}) \ldots \tan(A_{i_{2k+1}})$$

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