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.
 A: 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.
A: You only need to work modulo $4$, because
 $\sin\left(\dfrac{n\pi}2\right)$ is $0$ is $n$ is even, (i.e. $n\equiv 0,2 \mod 4)$) 
$1$ if $n=4k+1$ for some $k\in\Bbb{Z}$, (i.e. $n\equiv 1 \mod 4)$)
$-1$ if $n=4l+3$ for some $l\in\Bbb{Z}$ (i.e. $n\equiv 3 \mod 4)$)
Now, you only need to check that  if any of $a$ or $b$ is even, then the LHS of your identity is $0$, since $n=ab$ is even, and so is the RHS.
So, we have to check only the cases, when they are both odd.
If you multiply two odd numbers of the form $4k+1$, you get  $a.b=(4k+1).(4l+1)=4(4kl+k+l)+1$ which is of the form $4r+1$, so this corresponds to the LHS of your identity being $1\times 1$ and the RHS being $1$.
If you multiply two odd numbers of the form $4k+3$, you get  $a.b=(4k+3).(4l+3)=4(4kl+k+2)+1$ which is of the form $4r+1$, so this corresponds to the LHS of your identity being $(-1)\times (-1)$ and the RHS being $1$.
If you multiply two odd numbers of the form $4k+1,4l+3$, you get  $a.b=(4k+1).(4l+3)=4(4kl+3k+l)+3$ which is of the form $4r+3$, so this corresponds to the LHS of your identity being $1\times (-1)$ and the RHS being $-1$.
The last three paragraphs are concise if the modulo notation is used.
A: 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

$$\implies \sin\frac{(2A+1)\pi}{2}\,\sin\frac{(2B+1)\pi}{2}=\sin \frac{(2A+1)(2B+1)\pi}{2}=$$
$$\sin \frac{(4AB+2A+2B+1)\pi}{2}=\sin \left(\frac \pi 2+k\pi\right)=\pm1$$
A: You have that $\sin (k\frac {\pi} 2)$ is $0$ if $k$ is even, $1$ if $k\equiv 1 \mod 4$ or $- 1$ if $k\equiv -1 \mod 4$. Now you have that if at least one between $a$ and $b$ is even, it is too $ab$ and the equality olds. If $a\equiv b\equiv 1 \mod 4$ then $ab\equiv 1\mod 4$ and equality olds. The cases $a\equiv b\equiv - 1\mod 4$ and $a\equiv 1\equiv - b \mod 4$ are totally similar and cover all the possible cases for $a$ and $b$.
