<s>[INST] Proving $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin\frac{ab\pi}{2}$ for $a,b\in\mathbb{N}$? I was playing with the graphs of $f:\,\mathbb{R}\to\mathbb{R},\, f(x)=\sin axt\pi$ and $g:\,\mathbb{R}\to\mathbb{R},\, g(x)=\sin xt\pi\sin at\pi$.
What I conjectured is the following:

If $t=\frac{m}{2}$ where $m$ is $1$ above a multiple of $4$ (or if $m$ is even), then $f=g$ for all $a,x\in\mathbb{N}$.

I'm the most interested in the simplest non-trivial case, which is $m=1$:
$$\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin \frac{ab\pi}{2}\quad a,b\in\mathbb{N}.$$
If this is true, how could it be rigorously proved?
I tried rewriting $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}$ as
$$\frac{1}{4}\left(e^{\frac{\pi i}{2}(a-b)}+e^{-\frac{\pi i}{2}(a-b)}-e^{\frac{\pi i}{2}(a+b)}-e^{-\frac{\pi i}{2}(a+b)}\right)$$
and rewriting $\sin\frac{ab\pi}{2}$ as
$$\frac{1}{2i}\left(e^{\frac{ab\pi i}{2}}-e^{-\frac{ab\pi i}{2}}\right),$$
but that doesn't seem to help when I want to couple $a$ and $b$ so that there is a product of them in both expressions.
 [/INST] We have that

*

*$a\lor b$ even

$$\implies \sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin \frac{ab\pi}{2}=\sin (k\pi)=0$$

*

*$a\land b$ odd

$$\implies \sin\frac{(2A+1)\pi}{2}\,\sin\frac{(2B+1)\pi}{2}=\sin \frac{(2A+1)(2B+1)\pi}{2}=$$
$$\sin \frac{(4AB+2A+2B+1)\pi}{2}=\sin \left(\frac \pi 2+k\pi\right)=\pm1$$
</s>