三角形 $ABC$ 满足 $AB = 3$，$AC = 4$ 且角 $BAC = \theta$ 弧度。 三角形 $ABC$ 满足 $AB = 3$，$AC = 4$ 且角 $BAC = \theta$ 弧度。

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

$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos\theta$
$BC^2 = 3^2 + 4^2 - 2(3)(4)\cos\theta$
$BC^2 = 25 - 24\cos\theta$

2. Apply the small angle approximation for $\cos\theta$ to show the first required result.

Since $\theta$ is sufficiently small, we use the approximation $\cos\theta \approx 1 - \frac{\theta^2}{2}$.
$BC^2 \approx 25 - 24\left(1 - \frac{\theta^2}{2}\right)$
$BC^2 \approx 25 - 24 + 12\theta^2$
$BC^2 \approx 1 + 12\theta^2$
$BC \approx \sqrt{1 + 12\theta^2} \quad \text{(shown)}$

3. Expand $(1 + 12\theta^2)^{\frac{1}{2}}$ using the standard binomial series expansion up to the $\theta^4$ term.

$BC \approx (1 + 12\theta^2)^{\frac{1}{2}}$
Using $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2!}x^2$,
$BC \approx 1 + \left(\frac{1}{2}\right)(12\theta^2) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(12\theta^2)^2$

4. Simplify the expansion to determine the constants $p$ and $q$.

$BC \approx 1 + 6\theta^2 - \frac{1}{8}(144\theta^4)$
$BC \approx 1 + 6\theta^2 - 18\theta^4$
Comparing this with the given form $1 + p\theta^2 + q\theta^4$, we obtain:
$p = 6$
$q = -18$

1. Recall the polynomial approximation derived in part (i).

From part (i), expanding using the standard binomial series yields:
$(1 + 12\theta^2)^{\frac{1}{2}} \approx 1 + \frac{1}{2}(12\theta^2) + \frac{\frac{1}{2}(-\frac{1}{2})}{2!}(12\theta^2)^2 = 1 + 6\theta^2 - 18\theta^4$

2. Substitute $\theta = \frac{1}{4}$ into the radical expression to relate it to $\sqrt{7}$.

$\sqrt{1 + 12\left(\frac{1}{4}\right)^2} = \sqrt{1 + 12\left(\frac{1}{16}\right)} = \sqrt{1 + \frac{3}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$

3. Substitute $\theta = \frac{1}{4}$ into the polynomial approximation and equate to find $\sqrt{7}$.

$1 + 6\left(\frac{1}{4}\right)^2 - 18\left(\frac{1}{4}\right)^4 = 1 + \frac{6}{16} - \frac{18}{256} = 1 + \frac{3}{8} - \frac{9}{128}$
$= \frac{128}{128} + \frac{48}{128} - \frac{9}{128} = \frac{167}{128}$
Equating the two expressions:
$\frac{\sqrt{7}}{2} \approx \frac{167}{128}$
$\sqrt{7} \approx \frac{167}{64}$

1. State the conclusion and provide a valid mathematical reason referencing the size of theta.

Yes, it would be a better estimate. Since $\frac{1}{10}$ is closer to $0$ than $\frac{1}{4}$, the values of the higher-order terms that were omitted from the expansion (such as $\theta^6, \theta^8$, etc.) are much smaller. Thus, the truncation error is reduced and the polynomial approximation is more accurate.