证明 $\frac{\text{d}}{\text{d}\theta}(\cos^4 \theta + \sin^4 \theta) = p \sin q\theta$，其中 $p$ 和 $q$ 是整数。 由此求出对于 $0 < \theta < \frac{\pi}{2}$，曲线 $y = \cos^4 \theta + \sin^4 \theta$ 具有水平切线时 $\theta$ 的确切值。

1. Differentiate the expression with respect to $\theta$ using the chain rule

$\frac{\text{d}}{\text{d}\theta}(\cos^4 \theta + \sin^4 \theta) = 4\cos^3 \theta (-\sin \theta) + 4\sin^3 \theta (\cos \theta)$

2. Factorise the resulting expression

$= -4\sin\theta \cos^3\theta + 4\sin^3\theta \cos\theta$
$= -4\sin\theta \cos\theta (\cos^2\theta - \sin^2\theta)$

3. Apply double angle identities for sine and cosine

Since $2\sin\theta \cos\theta = \sin 2\theta$ and $\cos^2\theta - \sin^2\theta = \cos 2\theta$,
$= -2(2\sin\theta \cos\theta)(\cos^2\theta - \sin^2\theta)$
$= -2\sin 2\theta \cos 2\theta$

4. Apply the double angle identity for sine again to reach the final form

Since $2\sin 2\theta \cos 2\theta = \sin 4\theta$,
$= -\sin 4\theta$
(Thus $p = -1$ and $q = 4$)

1. Recall the derivative from part (i) and set it to zero

From part (i), $\frac{\text{d}}{\text{d}\theta}(\cos^4 \theta + \sin^4 \theta) = -\sin 4\theta$.
For a horizontal tangent, $\frac{\text{d}y}{\text{d}\theta} = 0$.
$-\sin 4\theta = 0$

2. Solve the trigonometric equation for the given domain

$\sin 4\theta = 0$
Since $0 < \theta < \frac{\pi}{2}$, the domain for $4\theta$ is $0 < 4\theta < 2\pi$.
The only solution in this range is $4\theta = \pi$.
$\theta = \frac{\pi}{4}$