We know generally: $$\mathcal{F}\left(f(t-t_{0})\right)=e^{-i\omega t_{0}}\mathcal{F}\left(f(t)\right)$$
And:
$$\mathcal{F}\left(cos(at)\right) = \sqrt{\frac{\pi}{2}}\left(\delta(\omega-a)+\delta(\omega+a)\right)$$
We can calculate FFT of $sin(ax)$:
$$\mathcal{F}\left(cos(at-\frac{\pi}{2})\right)=e^{-i\omega\frac{\pi}{2}}\sqrt{\frac{\pi}{2}}\left(\delta(\omega-a)+\delta(\omega+a)\right)$$
But it is not generally the FFT of $sin(ax)$. It is (calculating straight from the definition:
$$\sqrt{\frac{\pi}{2}}\left(\delta(\omega-a)e^{-i\frac{\pi}{2}}+\delta(\omega+a)e^{i\frac{\pi}{2}}\right)$$
Are both results similar? I fail to see it...
You've calculated the variable shift incorrectly: if $f(x)=\cos(ax)$, then $f(x-b) = \cos(ax-ab)$. So in fact $$ \mathcal{F}( \cos(a(x-\pi/(2a))) ) = e^{-i(\pi/2)(\omega/a)} \sqrt{\frac{\pi}{2}} (\delta(\omega-a) + \delta(\omega+a)). $$ But by the defining property of $\delta$, the exponential on the first term is the same as $e^{-i\pi/2}$, that on the second term the same as $e^{i\pi/2}$, so it comes out as the same as calculating $\mathcal{F}(\sin(ax))$ directly.