# General formula for sin &#x2061;<!-- ⁡ --> ( A 1 </msub> + A 2 </

General formula for $\mathrm{sin}\left({A}_{1}+{A}_{2}+\cdots +{A}_{n}\right)$
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
$\mathrm{sin}\left({A}_{1}+{A}_{2}+\cdots +{A}_{n}\right)=\mathrm{cos}{A}_{1}\mathrm{cos}{A}_{1}\cdots \mathrm{cos}{A}_{n}\left(\sum _{i=1}^{n}\mathrm{tan}{A}_{i}-\sum _{1\le i
And I cannot prove it. Can anyone help me prove or disprove it?
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Immanuel Glenn
It's not true for n=5:
$\left(\text{left side}\right)-\left(\text{right side}\right)=\mathrm{sin}\left({A}_{1}\right)\mathrm{sin}\left({A}_{2}\right)\mathrm{sin}\left({A}_{3}\right)\mathrm{sin}\left({A}_{4}\right)\mathrm{sin}\left({A}_{5}\right)$
I suspect that on the right side, for each k with $2k+1\le n$ you need
$\left(-1{\right)}^{k}\sum _{1\le {i}_{1}<\dots <{i}_{2k+1}}\mathrm{tan}\left({A}_{{i}_{1}}\right)\dots \mathrm{tan}\left({A}_{{i}_{2k+1}}\right)$