Hint needed to show that $a\cos t+b\sin t\leq \sqrt{a^2+b^2}$ for $t\in [0,2\pi)$ And that the upper bound is achieved for some choice of $\theta$. This exercise shows up in the Cauchy-Schwarz section of a textbook I am looking through but I don't see how to apply CS to prove. I would prefer a hint towards how to use this ineq. specifically.

Through standard techniques, you can see that the maximum of 
$$
f(t)=a\cos t+b\sin t-\sqrt{a^2+b^2}
$$
occurs for $\arctan\frac{a}{b}$ provided $t\ne \frac{\pi}{2},\frac{3\pi}{2}$.  Not sure if I can do much from there though.
Simply imagine that $(a, b)$ and $(\cos{t}, \sin{t})$ are two vectors, $a\cos{t} + b\sin{t}$ being the dot product of them.
Cauchy Schwartz on an arbitrary point and a point on the unit circle.