jmc

By the Cauchy-Schwarz Inequality, $$(a\cos \theta +b\sin \theta {)}^2\le (a^2+b^2)({\cos }^2\theta +{\sin }^2\theta )=a^2+b^2,$$so $a\cos \theta +b\sin \theta \le \sqrt{a^2+b^2}.$

If $a=b=0,$ then $a\cos \theta +b\sin \theta =0$ for all $\theta .$ Otherwise, $a^2+b^2>0,$ and we can find an angle $\theta$ such that $$\cos \theta =\frac{a}{\sqrt{a^2+b^2}}\text{and}\sin \theta =\frac{b}{\sqrt{a^2+b^2}},$$which makes $a\cos \theta +b\sin \theta =\sqrt{a^2+b^2}.$ Thus, the maximum value is $\boxed{\sqrt{a^2+b^2}}.$