$$
\text { Three. Given } f(\theta)=a \sin \theta+b \cos \theta \text {, }
$$

if $\mathrm{f}\left(\frac{\pi}{6}\right)=\sqrt{3}$, and the maximum value of $\mathrm{f}(\theta)$

is 2, find the value of $\mathrm{f}\left(\frac{\pi}{3}\right)$.

Three. Solution
Given $f\left(\frac{\pi}{6}\right)=\sqrt{3}$, we have
$$
\operatorname{asin} \frac{\pi}{6}+b \cos \frac{\pi}{6}=\sqrt{3},
$$

which simplifies to $a+\sqrt{3} b=2 \sqrt{3}$.
Also, $\mathrm{f}(\theta)=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}} \sin (\theta+\varphi)$, with the maximum value being $\sqrt{\mathrm{a}^{2}+b^{2}}$, so we have
$$
\left\{\begin{array}{l}
a+\sqrt{3} b=2 \sqrt{3}, \\
a^{2}+b^{2}=4
\end{array}\right.
$$

Solving these equations, we get $\left\{\begin{array}{l}a=\sqrt{3}, \\ b=1\end{array}\right.$ or $\left\{\begin{array}{l}a=0, \\ b=2\end{array}\right.$
$$
\therefore \mathrm{f}\left(\frac{\pi}{3}\right)=\sqrt{3} \sin \frac{\pi}{3}+\mathrm{c} \text { os } \frac{\pi}{3}=2 \text {, }
$$

or $\mathrm{f}\left(\frac{\pi}{3}\right)=0 \sin \frac{\pi}{3}+2 \cos \frac{\pi}{3}=1$.