在三角形 $ABC$ 中，角 $BAC = \theta$ 弧度，角 $ACB = \frac{\pi}{6}$ 弧度，且 $AC = \sqrt{3}$。 已知 $\theta$ 足够小，证明： $AB \approx \frac{2\sqrt{3}}{2 + 2\sqrt{3}\theta - \theta^2} \approx a + b\theta + c\theta^2,$ 其中 $a$，$b$ 和 $c$ 是待确定的常数，其值需用精确形式表示。

1. Set up the Sine Rule to express $AB$ in terms of $\theta$.

Let the third angle be $\angle ABC$.
Since angles in a triangle sum to $\pi$ radians:
$\angle ABC = \pi - \left(\theta + \frac{\pi}{6}\right)$
Using the Sine Rule in $\triangle ABC$:
$\frac{AB}{\sin\left(\frac{\pi}{6}\right)} = \frac{AC}{\sin(\angle ABC)}$
$\frac{AB}{\frac{1}{2}} = \frac{\sqrt{3}}{\sin\left(\pi - \left(\theta + \frac{\pi}{6}\right)\right)}$

2. Simplify the denominator using trigonometric identities.

Using the property $\sin(\pi - x) = \sin x$:
$AB = \frac{\frac{\sqrt{3}}{2}}{\sin\left(\theta + \frac{\pi}{6}\right)}$
Applying the compound angle formula for sine:
$\sin\left(\theta + \frac{\pi}{6}\right) = \sin\theta \cos\frac{\pi}{6} + \cos\theta \sin\frac{\pi}{6}$
$= \frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta$
Substituting this back into the expression for $AB$:
$AB = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta} = \frac{\sqrt{3}}{\sqrt{3}\sin\theta + \cos\theta}$

3. Substitute small angle approximations to derive the first required expression.

For sufficiently small $\theta$, we use the approximations $\sin\theta \approx \theta$ and $\cos\theta \approx 1 - \frac{\theta^2}{2}$:
$AB \approx \frac{\sqrt{3}}{\sqrt{3}\theta + \left(1 - \frac{\theta^2}{2}\right)}$
Multiplying the numerator and denominator by $2$ to clear the fractions:
$AB \approx \frac{2\sqrt{3}}{2\sqrt{3}\theta + 2 - \theta^2}$
Rearranging the denominator gives the first required expression:
$AB \approx \frac{2\sqrt{3}}{2 + 2\sqrt{3}\theta - \theta^2} \quad \text{(shown)}$

4. Prepare the expression for a binomial expansion.

To express this in the form $a + b\theta + c\theta^2$, factor out $2$ from the denominator to set up a $(1+x)^n$ binomial expansion:
$AB \approx \frac{2\sqrt{3}}{2 \left(1 + \sqrt{3}\theta - \frac{\theta^2}{2}\right)}$
$= \sqrt{3} \left(1 + \left(\sqrt{3}\theta - \frac{\theta^2}{2}\right)\right)^{-1}$

5. Apply the binomial expansion up to the $\theta^2$ term.

Using the binomial expansion $(1+x)^{-1} \approx 1 - x + x^2$ for small $x$, where $x = \sqrt{3}\theta - \frac{\theta^2}{2}$:
$AB \approx \sqrt{3} \left[ 1 - \left(\sqrt{3}\theta - \frac{\theta^2}{2}\right) + \left(\sqrt{3}\theta - \frac{\theta^2}{2}\right)^2 \right]$
Expanding and discarding terms with powers of $\theta$ greater than $2$:
$AB \approx \sqrt{3} \left[ 1 - \sqrt{3}\theta + \frac{\theta^2}{2} + \left(\sqrt{3}\theta\right)^2 \right]$
$AB \approx \sqrt{3} \left( 1 - \sqrt{3}\theta + \frac{\theta^2}{2} + 3\theta^2 \right)$
$AB \approx \sqrt{3} \left( 1 - \sqrt{3}\theta + \frac{7}{2}\theta^2 \right)$

6. Multiply out to find the exact values of the constants $a$, $b$, and $c$.

Multiplying the scalar $\sqrt{3}$ through the brackets:
$AB \approx \sqrt{3} - 3\theta + \frac{7\sqrt{3}}{2}\theta^2$
By comparing this sequence of terms to $a + b\theta + c\theta^2$, we find the exact values of the constants:
$a = \sqrt{3}, \quad b = -3, \quad c = \frac{7\sqrt{3}}{2}$