I've seen that arccos has $D_f [-1,1]$ however when putting arccos(2) into wolframalpha, I get the complex number result if I understand it correctly. Why is it so?
If true, should not $D_f$ be differently precised?
2026-04-20 10:43:16.1776681796
Why arccos(2) have value in complex numbers even though its $D_f [-1,1]$
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If you are working within $\mathbb R$ then $arcos (2)$ does not exist. But in the complex plane there are infinitely many solutions of the equation $\cos(z)=2$.