已知 $\theta$ 很小，证明 $\sec\theta \approx 1 + \tfrac{1}{2}\theta^{2}$。 下图显示了一个圆，圆心为 O，半径为 r，圆周上有两点 A 和 B，使得 $\angle AOB = \theta$ 弧度，其中 $\theta$ 很小。AC 是圆在点 A 处的切线，而 OBC 是一条直线。 证明弦 AB 的长度可以近似为 $r\theta$。 由此证明三角形 ABC 的周长可以近似为 $r\theta(a + b\theta)$，其中 $a$ 和 $b$ 是待确定的常数。

1. Express $\sec\theta$ in terms of $\cos\theta$ and apply the small angle approximation.

$\sec \theta = \frac{1}{\cos \theta}$
For small $\theta$, $\cos \theta \approx 1 - \frac{1}{2}\theta^2$.
Therefore, $\sec \theta \approx \frac{1}{1 - \frac{1}{2}\theta^2} = \left(1 - \frac{1}{2}\theta^2\right)^{-1}$.

2. Expand using the binomial series to obtain the final approximation.

Using the standard binomial expansion $(1+x)^n \approx 1+nx$ for small $x$, we expand up to the $\theta^2$ term:
$\left(1 - \frac{1}{2}\theta^2\right)^{-1} \approx 1 + (-1)\left(-\frac{1}{2}\theta^2\right)$
$\sec \theta \approx 1 + \frac{1}{2}\theta^2$

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

$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos\theta$
$AB^2 = r^2 + r^2 - 2(r)(r)\cos\theta$
$AB^2 = 2r^2 - 2r^2\cos\theta$
$AB^2 = 2r^2(1 - \cos\theta)$

2. Apply the small angle approximation for $\cos\theta$.

Since $\theta$ is small, we use the standard approximation $\cos\theta \approx 1 - \frac{1}{2}\theta^2$:
$AB^2 \approx 2r^2\left[1 - \left(1 - \frac{1}{2}\theta^2\right)\right]$
$AB^2 \approx 2r^2\left(\frac{1}{2}\theta^2\right)$
$AB^2 \approx r^2\theta^2$

3. Take the square root to show the required approximation for $AB$.

$AB \approx \sqrt{r^2\theta^2}$
$AB \approx r\theta \quad \text{(shown)}$

1. Find the length of $AC$ and apply the small angle approximation for tangent.

Since $AC$ is the tangent to the circle at $A$, $\angle OAC = 90^\circ$.
In right-angled $\triangle OAC$:
$AC = r \tan \theta$
Using the small angle approximation $\tan \theta \approx \theta$:
$AC \approx r\theta$

2. Express the length of $BC$ using $OC$ and the radius $OB$.

In $\triangle OAC$, $OC = \frac{r}{\cos \theta} = r \sec \theta$.
Since $O$, $B$, and $C$ are collinear and $B$ lies on the circumference, $OB = r$.
The length of $BC$ is:
$BC = OC - OB = r \sec \theta - r$

3. Apply the secant approximation from part (a) to find an expression for $BC$.

From part (a), $\sec \theta \approx 1 + \frac{1}{2}\theta^2$.
$BC \approx r\left(1 + \frac{1}{2}\theta^2\right) - r$
$BC \approx r + \frac{1}{2}r\theta^2 - r$
$BC \approx \frac{1}{2}r\theta^2$

4. Calculate the perimeter of $\triangle ABC$ and determine constants $a$ and $b$.

The perimeter of $\triangle ABC$ is the sum of its sides:
$Perimeter = AB + BC + AC$
Using the approximation for chord $AB \approx r\theta$ from part (b)(i):
$Perimeter \approx r\theta + \frac{1}{2}r\theta^2 + r\theta$
$Perimeter \approx 2r\theta + \frac{1}{2}r\theta^2$
Factorising out $r\theta$ to match the required form:
$Perimeter \approx r\theta\left(2 + \frac{1}{2}\theta\right)$
Comparing this with $r\theta(a + b\theta)$, we find:
$a = 2$, $b = \frac{1}{2}$