Explaining the trigonometric addition formulas via composition of rotations

Question

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.

Answer 1

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:

$\cos(\alpha+\beta)+i\sin(\alpha+\beta)=\exp(i(\alpha+\beta))=\exp(i\alpha)\cdot\exp(i\beta)=$
$=(\cos(\alpha)+i\sin(\alpha))\cdot(\cos(\beta)+i\sin(\beta))=$
$=(\cos(\alpha)\cdot\cos(\beta)-\sin(\alpha)\cdot\sin(\beta))+i(\cos(\alpha)\cdot\sin(\beta)+\sin(\alpha)\cdot\cos(\beta))$

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

--- rk

Answer 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.