In High School, the above mentioned formula is often proved assuming that $x,y$ are acute angles and using some more or less 'brilliant' geometrical construction.
And after that, books 'forget' the assumption about the angles and apply the identity to whatever angles.
There is a trivial, long and tedious proof by cases, with all combinations of the possible quadrants where $x$ and $y$ can lie.
Is there a shorter way to extend the formula for whatever angles? The proof must be suitable for High School, that is, no Taylor series or Euler's formula.
I don't think that this is a duplicate of the proposed question, because the first answer considers only acute angles and the second uses Euler's formula.
Best Answer
I learned this proof from this wonderful 1-page paper Proof of Sum and Difference Identities by Gilles Cazelais
$$ d = \sqrt{(\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2} \\ d= \sqrt{(\cos(\alpha-\beta)-1)^2+(\sin(\alpha-\beta)-0)^2} $$
from which you could even leave as an exercise for high school students to prove.