已知向量 $\mathbf{a}$ 和 $\mathbf{b}$ 为 $\mathbf{a}=\mathbf{i}+(\sin\theta)\mathbf{j}+(\cos\theta)\mathbf{k}\quad$ 且 $\quad\mathbf{b}=\mathbf{i}+(\sin\phi)\mathbf{j}+(\cos\phi)\mathbf{k}$，其中 $0\le\theta\le\phi\le\pi$。

1. Find the cross product $\mathbf{a}\times\mathbf{b}$.

$\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & \sin\theta & \cos\theta \\ 1 & \sin\phi & \cos\phi \end{vmatrix}$
$= (\sin\theta\cos\phi - \cos\theta\sin\phi)\mathbf{i} - (\cos\phi - \cos\theta)\mathbf{j} + (\sin\phi - \sin\theta)\mathbf{k}$
$= \sin(\theta-\phi)\mathbf{i} + (\cos\theta - \cos\phi)\mathbf{j} + (\sin\phi - \sin\theta)\mathbf{k}$

2. Set up the square of the magnitude $|\mathbf{a}\times\mathbf{b}|^2$.

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

3. Expand the squared terms and group trigonometric identities.

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

4. Simplify the expression using the compound angle formula for cosine.

$|\mathbf{a}\times\mathbf{b}|^2 = \sin^2(\theta-\phi) + 1 + 1 - 2\cos(\theta-\phi)$
$= \sin^2(\theta-\phi) + 2 - 2\cos(\theta-\phi)$

5. Substitute $\theta-\phi = -2\delta$ and express entirely in terms of $\delta$ using double angle formulas.

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

6. Factorise and evaluate the square root to find $|\mathbf{a}\times\mathbf{b}|$.

$|\mathbf{a}\times\mathbf{b}|^2 = 4\sin^2\delta\cos^2\delta + 4\sin^2\delta$
$= 4\sin^2\delta(\cos^2\delta + 1)$
Taking the positive square root:
$|\mathbf{a}\times\mathbf{b}| = \sqrt{4\sin^2\delta(1+\cos^2\delta)} = 2|\sin\delta|\sqrt{1+\cos^2\delta}$
Since $0 \le \theta \le \phi \le \pi$, we have $0 \le \phi-\theta \le \pi$, which means $0 \le \delta \le \frac{\pi}{2}$.
In this range, $\sin\delta \ge 0$, so $|\sin\delta| = \sin\delta$.
$|\mathbf{a}\times\mathbf{b}| = 2\sin\delta\sqrt{1+\cos^2\delta}$

1. Retrieve the expression for $|\mathbf{a}\times\mathbf{b}|$ from part (a)

$\mathbf{a}\times\mathbf{b} = \begin{pmatrix} 1 \\ \sin\theta \\ \cos\theta \end{pmatrix} \times \begin{pmatrix} 1 \\ \sin\phi \\ \cos\phi \end{pmatrix} = \begin{pmatrix} \sin\theta\cos\phi - \cos\theta\sin\phi \\ \cos\theta - \cos\phi \\ \sin\phi - \sin\theta \end{pmatrix}$
$= \begin{pmatrix} -\sin(\phi-\theta) \\ 2\sin\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right) \\ 2\cos\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right) \end{pmatrix} = \begin{pmatrix} -\sin(2\delta) \\ 2\sin\left(\frac{\phi+\theta}{2}\right)\sin\delta \\ 2\cos\left(\frac{\phi+\theta}{2}\right)\sin\delta \end{pmatrix}$
$|\mathbf{a}\times\mathbf{b}|^2 = \sin^2(2\delta) + 4\sin^2\delta \left[ \sin^2\left(\frac{\phi+\theta}{2}\right) + \cos^2\left(\frac{\phi+\theta}{2}\right) \right]$
$= (2\sin\delta\cos\delta)^2 + 4\sin^2\delta = 4\sin^2\delta(\cos^2\delta + 1)$
$|\mathbf{a}\times\mathbf{b}| = 2\sin\delta\sqrt{1+\cos^2\delta}$

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. Deduce the expression for $\sin\alpha$ using the geometric property of the cross product

The magnitude of the cross product is related to the angle $\alpha$ between the vectors by:
$|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\alpha$
Substituting the expressions from the previous steps:
$2\sin\delta\sqrt{1+\cos^2\delta} = (\sqrt{2})(\sqrt{2})\sin\alpha$
$2\sin\delta\sqrt{1+\cos^2\delta} = 2\sin\alpha$
$\sin\alpha = \sin\delta\sqrt{1+\cos^2\delta} \quad \text{(shown)}$