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.

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    $\begingroup$ I really don't think that it is a duplicate. The proposed question is not the same, and its answers don't answer mine. $\endgroup$
    – ajotatxe
    Commented Nov 21, 2019 at 15:14
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    $\begingroup$ @JoséCarlosSantos: The suggested duplicate is about a proof for acute angles. OP's question is specifically asking how to generalize the result to any angles. (I edited this question specifically so that the qualifier "for any angles $x$ and $y$" featured in the title.) $\endgroup$
    – Blue
    Commented Nov 21, 2019 at 15:18
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    $\begingroup$ Right. I have retracted my vote. $\endgroup$ Commented Nov 21, 2019 at 15:19
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    $\begingroup$ During my high school when everyone was shamelessly using it for obtuse angles as well I asked my teacher and she derived it assuming $\angle A$ as obtuse, thus $\pi-A$ will be acute. $\endgroup$
    – Naman Jain
    Commented Nov 21, 2019 at 16:51
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    $\begingroup$ Three close votes? Why on Earth ...?! $\endgroup$ Commented Nov 22, 2019 at 3:38

2 Answers 2


I learned this proof from this wonderful 1-page paper Proof of Sum and Difference Identities by Gilles Cazelais

Two black outlined circles with radius 1 is laid out side by side, each centered on x and y axes against a plain white background. Each circle has an outlined triangle inside with one corner in the center, and the other two on the outline of the circle of which the distance between is labelled d. On the left, both edges of the triangle from the center do not coincide with any of the axes; the counterclockwise angle from the x-axis of one of those edges is labelled alpha and the other beta. On the right, one of the edges coincide with the x-axis, the other making a counterclockwise angle from the x-axis of alpha minus beta. The corners of the triangles meeting the outline of the circle have their coordinates labelled as well, which are written in terms of cos, sin and their respective angles.

$$ d = \sqrt{(\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2} \\ d= \sqrt{(\cos(\alpha-\beta)-1)^2+(\sin(\alpha-\beta)-0)^2} $$

from which you could even leave as an exercise for high school students to prove.

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    $\begingroup$ This is the proof I remember learning as a kid. It gives you the cosine angle difference formula, from which the remaining formulas can be derived knowing that cosine is even, sine is odd, and $\sin x = \cos(\pi/2-x).$ $\endgroup$ Commented Nov 22, 2019 at 3:51
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    $\begingroup$ Which program have you used to make the figures? $\endgroup$
    – ajotatxe
    Commented Nov 22, 2019 at 14:05

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.


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