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.

Suppose $a$ or $b$ is even. Without loss of generality, $a=2k$. Then
$$\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(\frac{2k\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(k\pi\right)\sin\left(\frac{b\pi}{2}\right)=0$$
and
$$\sin\left(\frac{ab\pi}{2}\right)= \sin\left(\frac{2kb\pi}{2}\right)=\sin\left(kb\pi\right)=0$$
Suppose $a$ and $b$ are both odd. Splitting into further cases, suppose
$$a\equiv b\equiv 1\ (\text{mod }4)$$
Then
$$a=4k+1\text{ and }b=4j+1$$
This implies
$$\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(\frac{(4k+1)\pi}{2}\right)\sin\left(\frac{(4j+1)\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right)=1$$
and
$$\sin\left(\frac{ab\pi}{2}\right)= \sin\left(\frac{(4k+1)(4j+1)\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)=1$$
The cases where
$$a\equiv b\equiv 3\ (\text{mod }4)$$
and
$$a\equiv 1\ (\text{mod }4)$$
$$b\equiv 3\ (\text{mod }4)$$
can be worked out in a similar manner.
