三角形 $ABC$ 满足 $AC = \sqrt{2}$，$BC = 4$ 且角 $ACB = \frac{1}{4}\pi + \theta$。已知 $heta$ 足够小，以至于可以忽略 $heta^3$ 及更高阶项，证明 $AB \approx \sqrt{10}\left[1 + a\theta + b\theta^2\right],$ 其中 $a$ 和 $b$ 为实常数。

1. Apply the cosine rule to triangle $ABC$ to find an expression for $AB^2$.

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

2. Use the addition formula for cosine to expand the trigonometric term.

$AB^2 = 18 - 8\sqrt{2}\left(\cos\frac{\pi}{4}\cos\theta - \sin\frac{\pi}{4}\sin\theta\right)$
$AB^2 = 18 - 8\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos\theta - \frac{1}{\sqrt{2}}\sin\theta\right)$
$AB^2 = 18 - 8(\cos\theta - \sin\theta)$

3. Apply small angle approximations for $\sin\theta$ and $\cos\theta$.

For small $\theta$, $\cos\theta \approx 1 - \frac{\theta^2}{2}$ and $\sin\theta \approx \theta$.
$AB^2 \approx 18 - 8\left(1 - \frac{\theta^2}{2} - \theta\right)$
$AB^2 \approx 18 - 8 + 4\theta^2 + 8\theta$
$AB^2 \approx 10 + 8\theta + 4\theta^2$
$AB^2 \approx 10\left(1 + \frac{4}{5}\theta + \frac{2}{5}\theta^2\right)$

4. Find $AB$ by taking the square root and applying the binomial series expansion up to the term in $\theta^2$.

$AB = \sqrt{10}\left(1 + \frac{4}{5}\theta + \frac{2}{5}\theta^2\right)^{\frac{1}{2}}$
Using the standard expansion $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$:
$AB \approx \sqrt{10}\left[ 1 + \frac{1}{2}\left(\frac{4}{5}\theta + \frac{2}{5}\theta^2\right) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}\left(\frac{4}{5}\theta\right)^2 \right]$
$AB \approx \sqrt{10}\left[ 1 + \frac{2}{5}\theta + \frac{1}{5}\theta^2 - \frac{1}{8}\left(\frac{16}{25}\theta^2\right) \right]$
$AB \approx \sqrt{10}\left[ 1 + \frac{2}{5}\theta + \frac{1}{5}\theta^2 - \frac{2}{25}\theta^2 \right]$
$AB \approx \sqrt{10}\left[ 1 + \frac{2}{5}\theta + \frac{3}{25}\theta^2 \right]$
(Thus $a = \frac{2}{5}$ and $b = \frac{3}{25}$)