I am generally dissatisfied with the way trigonometric addition formulas like $$ \cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) $$ are derived in high school textbooks. There are numerous proofs, some of which are short but unintuitive, some of which introduce unnecessary calculations, many with restrictions on $\alpha$, $\beta$ and $\alpha + \beta$, like having to lie between $0$ and $\frac{\pi}{2}$ radians.

To me, the addition formulas are simply a coordinatization of the observation that composing a rotation by $\alpha$ with a rotation by $\beta$ yields a rotation by $\alpha + \beta$. Hence in my opinion the proper way to prove such a formula would be to observe that $$ \begin{bmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \end{bmatrix} = \begin{bmatrix} \cos(\beta) & -\sin(\beta) \\ \sin(\beta) & \cos(\beta) \end{bmatrix} \cdot \begin{bmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{bmatrix} $$ then work out the product on the right to obtain the formulas by equating entries. However, such a proof is beyond the scope of a high-school textbook, as high schoolers - if they know about matrices at all - are rarely taught the link between composition of linear transformations and matrix multiplication.

Is there a way to salvage the essence of this proof - that the formulas are merely a way of expressing that a rotation by $\alpha+ \beta$ can be obtained by composing rotations by $\alpha$ and $\beta$ - without using linear algebra? Of course complex numbers are also out of the question.

  • $\begingroup$ Arguably with the lengths properly interpreted (in particular to discuss negative lengths correctly) this sort of diagram is the geometric version of the matrix proof. $\endgroup$
    – Chappers
    May 30, 2018 at 21:02
  • $\begingroup$ I suspect that one could also adapt the proof based on $\cos$ and $\sin$ being solutions of the linear differential equation $u''+u=0$ with the right initial conditions to be more obviously to do with rotations, although it's not immediately obvious how. $\endgroup$
    – Chappers
    May 30, 2018 at 21:06
  • $\begingroup$ Indeed such a diagram is helpful and relatively intuitive, but I personally feel like I should pause and go over all possible cases. What if $\alpha + \beta$ is a multiple of $\pi$? Indeed how does one interpret the negative lengths? I am still hoping for a proof which would more clearly connect the formula to a composition of transformations. Differential equations are, I'm afraid, also not part of the background of a high school student who still has to learn trigonometry. Also, aren't the addition formulas used in finding the derivative of the sine and cosine functions? $\endgroup$
    – Bib-lost
    May 30, 2018 at 21:08
  • $\begingroup$ It depends on how you define sine and cosine: if it's geometrical, then you have to work out how to define them for angles larger than $\pi/2$. You can define the as the solutions to $u''+u=0$ that satisfy $\sin{0}=0$, $\sin'{0}=1$ and $\sin'=\cos$, or similar, and the rest of the theory can be derived from there. Or you can use power series, of course, although that's probably the worst of all. $\endgroup$
    – Chappers
    May 30, 2018 at 21:44
  • $\begingroup$ @Bib-lost Would you be ok with a derivation of your formula, using the dot product and distance formula? $\endgroup$
    – imranfat
    May 30, 2018 at 21:51

2 Answers 2


Can we use vectors? I assume so, since rotation is a transformation in vectors.

The $\alpha$ rotation sends $a\vec{I}+b\vec{j}$ to $A\vec{I}+B\vec{j}$ with $A:=a\cos\alpha-b\sin\alpha,\,B:=a\sin\alpha+b\cos\alpha$. If you compute where three second rotation sends the latter vector, you can complete the proof. It's the same logic as in the matrix case, but you don't have to explain the theory thereof.

  • $\begingroup$ Right, good idea. So then one only needs to spend a bit of time on explaining why a rotation by $\alpha$ degrees sends a point $(x, y)$ to $(x\cos(\alpha) - y\sin(\alpha), x\sin(\alpha) + y\cos(\alpha))$. Will try to think of a good explanation for this, let me know if you find anything. $\endgroup$
    – Bib-lost
    May 31, 2018 at 9:26
  • $\begingroup$ @Bib-lost If you can convince them of linearity, you just need to draw a diagram showing where $\vec{i},\,\vec{j}$ each end up. $\endgroup$
    – J.G.
    May 31, 2018 at 10:12
  • $\begingroup$ Of course. Will try this on Monday, see if it makes it click for them. $\endgroup$
    – Bib-lost
    May 31, 2018 at 10:14

You might take refuge to complex numbers instead, by means of the Euler relation $\exp(i\phi)=\cos(\phi)+i\sin(\phi)$ and the fundamental property of the $\exp$ function:


And finally use therefrom the real resp. imaginary part separately.
This is how you'd get the trigonometric addition theorems.

--- rk

  • $\begingroup$ Thanks. This satisfies the condition of relying on the composition property, but there is no way I can explain this to high schoolers without a background in complex numbers. $\endgroup$
    – Bib-lost
    May 30, 2018 at 21:42
  • $\begingroup$ Sure you can. Take the Taylor series of $\exp$, of $\cos$ and of $\sin$, and you will get the Euler relation just by using the defining property of $i^2=-1$. - In fact, in Germany matrices as well as complex numbers are being tought at school (Gymnasium). $\endgroup$ May 30, 2018 at 21:47
  • $\begingroup$ @Dr.RichardKlitzing But do they also teach Taylor Series in High School? Taylor series of the trigonometric functions come from somewhere too. If you derive those, you need derivatives of them and for that, sumformulas come back (circular reasoning) $\endgroup$
    – imranfat
    May 30, 2018 at 21:48
  • $\begingroup$ At least they have lots of analysis. So all the therein being used derivations are known. Therefore it wouldn't be hard to teach them anyway, if those aren't in the current curriculum. $\endgroup$ May 30, 2018 at 21:55
  • $\begingroup$ In fact you just need the respective Maclaurin series here. - I can't see, where you need the sum formulas there. You just need that the derivative of $\sin$ is $\cos$, and the other way round with a minus sign. Then you are done. $\endgroup$ May 30, 2018 at 22:09

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