if the equation $$\cos{(a\sin{x})}=\sin{(b\cos{x})}$$ have no zero solution,then $a^2+b^2$ range of value
$A:[0,\dfrac{\pi}{4})$,$B: [0,\dfrac{\pi^2}{2})$,$C: [0,\dfrac{\pi^2}{4})$,$D:[0,\dfrac{\pi}{2})$
My try: if the equation have solution ,then $$\cos{(a\sin{x})}=\sin{(b\cos{x})}\Longrightarrow \sin{\left(\dfrac{\pi}{2}-a\sin{x}\right)}=\sin{(b\cos{x})}$$ then we have $$\dfrac{\pi}{2}=a\sin{x}+b\cos{x}=\sqrt{a^2+b^2}\sin{(x+\phi)}$$ so $$a^2+b^2\le\dfrac{\pi^2}{4}$$
But I can't sure find which is true?