在三角形 $PQR$ 中，角 $PQR = \frac{\pi}{3}$ 弧度，角 $PRQ = \frac{\pi}{3} - \theta$ 弧度，其中 $\theta$ 足够小，使得 $\theta^3$ 及更高次幂的 $\theta$ 可以忽略不计。 证明 $\frac{PR}{PQ} \approx 1 + \alpha\theta + \beta\theta^2$，其中 $\alpha$ 和 $\beta$ 是待确定的精确实常数。

1. Apply the Sine Rule to express the ratio of $PR$ to $PQ$.

By the Sine Rule:
$\frac{PR}{\sin\left(\frac{\pi}{3}\right)} = \frac{PQ}{\sin\left(\frac{\pi}{3} - \theta\right)}$
$\frac{PR}{PQ} = \frac{\sin\left(\frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{3} - \theta\right)}$

2. Expand the denominator using the compound angle formula.

$\sin\left(\frac{\pi}{3} - \theta\right) = \sin\left(\frac{\pi}{3}\right)\cos\theta - \cos\left(\frac{\pi}{3}\right)\sin\theta$
$= \frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta$

3. Substitute the small angle approximations for sine and cosine.

Since $\theta$ is small, $\sin\theta \approx \theta$ and $\cos\theta \approx 1 - \frac{\theta^2}{2}$.
$\sin\left(\frac{\pi}{3} - \theta\right) \approx \frac{\sqrt{3}}{2}\left(1 - \frac{\theta^2}{2}\right) - \frac{1}{2}\theta$
$= \frac{\sqrt{3}}{2} - \frac{1}{2}\theta - \frac{\sqrt{3}}{4}\theta^2$

4. Substitute the denominator back into the ratio and factorise to prepare for binomial expansion.

$\frac{PR}{PQ} \approx \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} - \frac{1}{2}\theta - \frac{\sqrt{3}}{4}\theta^2}$
Divide the numerator and denominator by $\frac{\sqrt{3}}{2}$:
$\frac{PR}{PQ} \approx \frac{1}{1 - \frac{1}{\sqrt{3}}\theta - \frac{1}{2}\theta^2}$
$= \left(1 - \left(\frac{1}{\sqrt{3}}\theta + \frac{1}{2}\theta^2\right)\right)^{-1}$

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

Using $(1 - x)^{-1} \approx 1 + x + x^2$ for small $x$:
$\frac{PR}{PQ} \approx 1 + \left(\frac{1}{\sqrt{3}}\theta + \frac{1}{2}\theta^2\right) + \left(\frac{1}{\sqrt{3}}\theta + \frac{1}{2}\theta^2\right)^2$
Expanding and ignoring terms with $\theta^3$ and higher powers:
$\frac{PR}{PQ} \approx 1 + \frac{1}{\sqrt{3}}\theta + \frac{1}{2}\theta^2 + \left(\frac{1}{\sqrt{3}}\theta\right)^2$
$= 1 + \frac{\sqrt{3}}{3}\theta + \frac{1}{2}\theta^2 + \frac{1}{3}\theta^2$
$= 1 + \frac{\sqrt{3}}{3}\theta + \frac{5}{6}\theta^2$
Comparing coefficients, we obtain $\alpha = \frac{\sqrt{3}}{3}$ and $\beta = \frac{5}{6}$.