21. By contradiction. Assume $a^{2}+b^{2} \geqslant \frac{\pi^{2}}{4}$, since $a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \sin (x+$ $\varphi)$, where $\varphi$ is a fixed real number depending only on $a, b$, such that $\cos \varphi=\frac{a}{\sqrt{a^{2}+b^{2}}}, \sin \varphi=$ $\frac{b}{\sqrt{a^{2}+b^{2}}}$.

Since $\sqrt{a^{2}+b^{2}} \geqslant \frac{\pi}{2}$, there exists a real number $x_{0}$ such that $\sqrt{a^{2}+b^{2}} \sin \left(x_{0}+\varphi\right)=\frac{\pi}{2}$, i.e., $a \sin x_{0}+b \cos x_{0}=\frac{\pi}{2}$

This implies $\cos \left(a \sin x_{0}\right)>\sin \left(b \cos x_{0}\right)$, which contradicts the assumption. Therefore, $a^{2}+b^{2}<\frac{\pi^{2}}{4}$.