Find the minimum value of $5\sin \theta +12\cos \theta$.

The expression is of the form $a\sin \theta +b\cos \theta$.
The minimum and maximum values of $a\sin \theta +b\cos \theta$ are given by $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$ respectively.

Here, $a=5$ and $b=12$.

$$\text{Minimum value}=-\sqrt{5^2+12^2}$$

$$\text{Minimum value}=-\sqrt{25+144}=-\sqrt{169}=-13$$