Is there a short, elementary way to show $\cos(x+y)=\cos x\cos y-\sin x\sin y$ for any angles $x$ and $y$? 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.
 A: 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.
A: Durell & Robson's proof seems satisfactory to me. It refers back to their Elementary Trigonometry, Part III, "The General Angle and Compound Angles", for proofs that:

\begin{align*}
\cos\left(A + \frac\pi2\right) & = -\sin A, \\
\sin\left(A + \frac\pi2\right) & = \cos A,
\end{align*}
  for values of $A$ of any magnitude, positive or negative.

Unfortunately, my copy of E.T. doesn't contain Part III.
[Added later: I found a PDF copy. It gives Euclid-style proofs, by congruent triangles, with figures for the cases where $A$ is in the second or third quadrant. The reader is asked to supply figures for the cases of the first and fourth quadrants.]
It still seems worth typing out their short proof of the addition formulae from p.123f. of Advanced Trigonometry (1930, Dover reprint 2003):

Let the directed lines $O\xi, OP, O\eta$ make angles $A, A+B, A + \frac\pi2$ with $Ox$; and let the projections of $P$ on $O\xi, O\eta$ be $N, M$. Suppose that $OP$ contains $l$ units of length.
The positions of $N, M$ on the directed lines $O\xi, O\eta$ are given by the directed numbers which measure $ON, OM$, and these are, by the definitions of the cosine and sine of the general angle, $l\cos B, l\sin B$.
$\therefore$ by [an earlier equation],
  \begin{align*}
\text{Projection of } ON \text{ on } Ox & = l\cos B \cdot\cos A, \\
\text{Projection of } OM \text{ on } Ox & = l\sin B \cdot\cos\left(A + \frac\pi2\right). \\
\text{Also the projection of } OP \text{ on } Ox & = l\cos(A+B).
\end{align*}
But the projection of $OP$ on $Ox$ is equal to the sum of the projections on $ON, NP$, i.e. to the sum of the projections of $ON, OM$ on $Ox$.
  $$
\therefore\ l\cos(A+B) = l\cos B\cos A + l\sin B \cdot\cos\left(A + \frac\pi2\right).
$$
  But $\cos\left(A + \frac\pi2\right) = -\sin A$, see E.T., pp. 199, 200;
  $$
\therefore\ \cos(A+B) = \cos A\cos B - \sin A\sin B.
$$
  Further, if the directed line $Oy$ makes $+\frac\pi2$ with $Ox$, the projections of $ON, OM, OP$ on $Oy$ are
  $$
l\cos B\sin A, \ l\sin B\sin\left(A + \frac\pi2\right), \ l\sin(A+B).
$$
$\therefore$ as before,
  $$
l\sin(A+B) = l\cos B\sin A + l\sin B \cdot\sin\left(A + \frac\pi2\right).
$$
  But $\sin\left(A + \frac\pi2\right) = \cos A$, see E.T., pp. 199, 200;
  $$
\therefore\ \sin(A+B) = \sin A\cos B + \cos A\sin B.
$$
  This proof holds good for values of $A$ and $B$ of any magnitude, positive or negative. Figs 59, 60 [omitted] show two possible cases; the reader should draw other figures (e.g. $A = 100^\circ$, $B = 50^\circ$ or $A = 220^\circ$, $B = 160^\circ$) and satisfy himself that the proof applies to them, without any modification.

