"Beats" via trig identity or something? Planning on talking about resonance in DE.
The solution to the IVP $$y''+y=\cos(t),\quad y(0)=y'(0)=0$$is $$y=\frac12 t\sin(t).$$Resonance, great.
Now what if the forcing function has almost the resonant frequency? If $\alpha^2\ne1$ the solution to $$y''+y=\cos(\alpha t),\quad y(0)=y'(0)=0$$is $$y=\frac1{1-\alpha^2}(\cos(\alpha t)-\cos(t)).$$It's pleasant to note that this is $\sim\frac12 t\sin(t)$ as $t\to0$, and that for every $t$ it tends to $\frac12 t\sin(t)$ as $\alpha\to1$.
For a fixed $\alpha$ close to $1$ it's clear if you think about it that $\cos(\alpha t)-\cos(t)$ exhibits "beats": there are intervals where it's close to $2\cos(\alpha t)$ and intervals where it's close to $0$. Alas when I say "you" here it's also clear that I mean you, not my DE students.
Hence the question: Is there a clever way to  write $\cos(\alpha t)-\cos(t)$ so that the beats are evident? Like it's clear that $\cos(\epsilon t)\cos(t)$ shows beats - something analogous to that? Of course $$e^{i\alpha t}-e^{it}=e^{it}(e^{i(\alpha-1)t}-1),$$but I'd rather avoid complex numbers here and sadly the real part of a product is not the product of the real parts. I could use that, or just the sum formula for the cosine, to get a sum of two products, each of which is the sort of thing I want. But writing it as just the product of something with frequency $1$ and something with low frequency would be so much nicer.
This must be well studied by people who study such things...

hint: $$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$
