【Example 4.23】Find $x=\tan 20^{\circ}+4 \sin 20^{\circ}$. (1993 College Entrance Examination Question)
$$
\begin{array}{l} 
\text { Sol } \text { Original expression }=\frac{\sin 20^{\circ}+2 \sin 40^{\circ}}{\cos 20^{\circ}}=\frac{\sin 20^{\circ}+2 \sin \left(60^{\circ}-20^{\circ}\right)}{\cos 20^{\circ}} \\
=\sqrt{3} .
\end{array}
$$

Sol Original expression $=\frac{\sin 20^{\circ}+2 \sin 40^{\circ}}{\cos 20^{\circ}}=\frac{\sin 20^{\circ}+2 \sin \left(60^{\circ}-20^{\circ}\right)}{\cos 20^{\circ}}$ $=\sqrt{3}$.

The idea is: Think of $60-20=40$, expand $\sin \left(60^{\circ}-20^{\circ}\right)=\sqrt{3} \cos 20^{\circ}$ $-\sin 20^{\circ}$, eliminate $\sin 20^{\circ}$, cancel out $\cos 20^{\circ}$ to get $x=\sqrt{3}$.