向量 $\mathbf{a}$ 和 $\mathbf{b}$ 分别由下式给出： $\mathbf{a} = \mathbf{i} + (\sin\theta)\mathbf{j} + (\cos\theta)\mathbf{k} \quad \text{且} \quad \mathbf{b} = \mathbf{i} + (\sin\phi)\mathbf{j} + (\cos\phi)\mathbf{k} \text{，其中 } 0 \le \theta \le \phi \le \pi .$

1. Compute the cross product of vectors $\mathbf{a}$ and $\mathbf{b}$ and simplify the x-component.

$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 1 \\ \sin\theta \\ \cos\theta \end{pmatrix} \times \begin{pmatrix} 1 \\ \sin\phi \\ \cos\phi \end{pmatrix}$
$= \begin{pmatrix} \sin\theta\cos\phi - \cos\theta\sin\phi \\ \cos\theta(1) - \cos\phi(1) \\ 1(\sin\phi) - 1(\sin\theta) \end{pmatrix}$
$= \begin{pmatrix} \sin(\theta - \phi) \\ \cos\theta - \cos\phi \\ \sin\phi - \sin\theta \end{pmatrix}$

2. Set up the equation for the square of the magnitude and expand the terms.

$|\mathbf{a} \times \mathbf{b}|^2 = \sin^2(\theta - \phi) + (\cos\theta - \cos\phi)^2 + (\sin\phi - \sin\theta)^2$
$= \sin^2(\theta - \phi) + \cos^2\theta - 2\cos\theta\cos\phi + \cos^2\phi + \sin^2\phi - 2\sin\theta\sin\phi + \sin^2\theta$

3. Group terms and apply Pythagorean and compound angle identities to simplify the expression.

Grouping the terms:
$|\mathbf{a} \times \mathbf{b}|^2 = \sin^2(\theta - \phi) + (\cos^2\theta + \sin^2\theta) + (\cos^2\phi + \sin^2\phi) - 2(\cos\theta\cos\phi + \sin\theta\sin\phi)$
Using $\cos^2 x + \sin^2 x = 1$ and the cosine difference identity $\cos(\theta - \phi) = \cos\theta\cos\phi + \sin\theta\sin\phi$:
$|\mathbf{a} \times \mathbf{b}|^2 = \sin^2(\theta - \phi) + 1 + 1 - 2\cos(\theta - \phi)$
Since $\sin^2(\theta - \phi) = \sin^2(\phi - \theta)$ and $\cos(\theta - \phi) = \cos(\phi - \theta)$:
$|\mathbf{a} \times \mathbf{b}|^2 = \sin^2(\phi - \theta) + 2 - 2\cos(\phi - \theta)$

4. Substitute $\delta = \frac{1}{2}(\phi - \theta)$ and factorise the expression completely using double-angle identities.

Given $\delta = \frac{1}{2}(\phi - \theta)$, we have $\phi - \theta = 2\delta$. Substituting this in:
$|\mathbf{a} \times \mathbf{b}|^2 = \sin^2(2\delta) + 2 - 2\cos(2\delta)$
Using the double-angle identities $\sin(2\delta) = 2\sin\delta\cos\delta$ and $\cos(2\delta) = 1 - 2\sin^2\delta$:
$|\mathbf{a} \times \mathbf{b}|^2 = (2\sin\delta\cos\delta)^2 + 2 - 2(1 - 2\sin^2\delta)$
$= 4\sin^2\delta\cos^2\delta + 4\sin^2\delta$
$= 4\sin^2\delta (\cos^2\delta + 1)$

5. Take the positive square root and justify the sign of $\sin\delta$ based on the given angle range.

Taking the square root gives:
$|\mathbf{a} \times \mathbf{b}| = 2|\sin\delta| \sqrt{1 + \cos^2\delta}$
We are given $0 \le \theta \le \phi \le \pi$, which means $0 \le \phi - \theta \le \pi$.
Since $\delta = \frac{1}{2}(\phi - \theta)$, it follows that $0 \le \delta \le \frac{\pi}{2}$.
In this principal range, $\sin\delta \ge 0$, so $|\sin\delta| = \sin\delta$.
$\therefore |\mathbf{a} \times \mathbf{b}| = 2\sin\delta \sqrt{1 + \cos^2\delta}$

1. Concisely re-derive the expression for $|\mathbf{a} \times \mathbf{b}|$ from part (a).

$\text{From part (a), we first determine } |\mathbf{a} \times \mathbf{b}|.$
$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 1 \\ \sin\theta \\ \cos\theta \end{pmatrix} \times \begin{pmatrix} 1 \\ \sin\phi \\ \cos\phi \end{pmatrix} = \begin{pmatrix} \sin\theta\cos\phi - \cos\theta\sin\phi \\ \cos\phi - \cos\theta \\ \sin\phi - \sin\theta \end{pmatrix}$
$\text{Using trigonometric factor formulae and } \phi - \theta = 2\delta:$
$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -\sin(2\delta) \\ -2\sin\left(\frac{\phi+\theta}{2}\right)\sin\delta \\ 2\cos\left(\frac{\phi+\theta}{2}\right)\sin\delta \end{pmatrix}$
$|\mathbf{a} \times \mathbf{b}|^2 = \sin^2(2\delta) + 4\sin^2\delta\left(\sin^2\left(\frac{\phi+\theta}{2}\right) + \cos^2\left(\frac{\phi+\theta}{2}\right)\right)$
$|\mathbf{a} \times \mathbf{b}|^2 = (2\sin\delta\cos\delta)^2 + 4\sin^2\delta(1) = 4\sin^2\delta(1+\cos^2\delta)$
$|\mathbf{a} \times \mathbf{b}| = 2\sin\delta\sqrt{1+\cos^2\delta}$

2. Calculate the magnitudes of vectors $\mathbf{a}$ and $\mathbf{b}$.

$|\mathbf{a}| = \sqrt{1^2 + \sin^2\theta + \cos^2\theta} = \sqrt{1 + 1} = \sqrt{2}$
$|\mathbf{b}| = \sqrt{1^2 + \sin^2\phi + \cos^2\phi} = \sqrt{1 + 1} = \sqrt{2}$

3. Relate the cross product magnitude to the angle $\alpha$ to deduce the given expression.

$\text{By definition, } |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\alpha.$
$\text{Substitute the values obtained:}$
$2\sin\delta\sqrt{1+\cos^2\delta} = (\sqrt{2})(\sqrt{2})\sin\alpha$
$2\sin\delta\sqrt{1+\cos^2\delta} = 2\sin\alpha$
$\sin\alpha = \sin\delta\sqrt{1+\cos^2\delta} \quad \text{(shown)}$