三角形 $ABC$ 满足 $AC = \sqrt{2}$，$BC = 4$ 且角 $ACB = \tfrac{1}{4}\pi + \theta$。已知 $\theta$ 足够小，以至于 $\theta^3$ 及更高阶的 $\theta$ 幂项可以忽略不计，证明： $AB \approx \sqrt{10}\bigl[1 + a\theta + b\theta^2\bigr],$ 其中 $a$ 和 $b$ 是实常数。

1. Use the cosine rule and the compound angle formula to find an exact expression for $AB^2$.

$AB^2 = AC^2 + BC^2 - 2(AC)(BC)\cos(\angle ACB)$
$AB^2 = (\sqrt{2})^2 + 4^2 - 2(\sqrt{2})(4)\cos\left(\frac{\pi}{4} + \theta\right)$
$AB^2 = 18 - 8\sqrt{2}\left(\cos\frac{\pi}{4}\cos\theta - \sin\frac{\pi}{4}\sin\theta\right)$
Since $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$AB^2 = 18 - 8\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos\theta - \frac{1}{\sqrt{2}}\sin\theta\right)$
$AB^2 = 18 - 8\cos\theta + 8\sin\theta$

2. Apply small angle approximations to simplify the expression for $AB^2$.

Given $\theta$ is small, $\cos\theta \approx 1 - \frac{\theta^2}{2}$ and $\sin\theta \approx \theta$.
$AB^2 \approx 18 - 8\left(1 - \frac{\theta^2}{2}\right) + 8(\theta)$
$AB^2 \approx 18 - 8 + 4\theta^2 + 8\theta$
$AB^2 \approx 10 + 8\theta + 4\theta^2$

3. Factorise the expression to prepare for the binomial expansion of $AB$.

$AB^2 \approx 10\left(1 + \frac{4}{5}\theta + \frac{2}{5}\theta^2\right)$
$AB \approx \sqrt{10}\left(1 + \frac{4}{5}\theta + \frac{2}{5}\theta^2\right)^{\frac{1}{2}}$

4. Expand using the binomial theorem up to the $\theta^2$ term.

Using the expansion $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$:
$AB \approx \sqrt{10}\left[ 1 + \frac{1}{2}\left(\frac{4}{5}\theta + \frac{2}{5}\theta^2\right) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}\left(\frac{4}{5}\theta\right)^2 \right]$
$AB \approx \sqrt{10}\left[ 1 + \frac{2}{5}\theta + \frac{1}{5}\theta^2 - \frac{1}{8}\left(\frac{16}{25}\theta^2\right) \right]$
$AB \approx \sqrt{10}\left[ 1 + \frac{2}{5}\theta + \frac{1}{5}\theta^2 - \frac{2}{25}\theta^2 \right]$
$AB \approx \sqrt{10}\left( 1 + \frac{2}{5}\theta + \frac{3}{25}\theta^2 \right)$
This is in the required form with $a = \frac{2}{5}$ and $b = \frac{3}{25}$.