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

1. Evaluate the cross product of vectors $\mathbf{a}$ and $\mathbf{b}$.

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

2. Write the expression for the squared magnitude $|\mathbf{a} \times \mathbf{b}|^2$ and expand the terms.

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

3. Simplify the expanded expression using Pythagorean and compound angle identities.

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

4. Substitute $\theta-\phi = -2\delta$ and apply 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 $\sin(2\delta) = 2\sin\delta\cos\delta$ and $\cos(2\delta) = 1 - 2\sin^2\delta$:
$= (2\sin\delta\cos\delta)^2 + 2 - 2(1 - 2\sin^2\delta)$
$= 4\sin^2\delta\cos^2\delta + 4\sin^2\delta$

5. Factorise the expression and evaluate the magnitude by taking the square root.

$|\mathbf{a} \times \mathbf{b}|^2 = 4\sin^2\delta(\cos^2\delta + 1)$
Taking the 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 $0 \le \phi-\theta \le \pi$, meaning $0 \le \delta \le \frac{\pi}{2}$.
Thus, $\sin\delta \ge 0$, giving:
$|\mathbf{a} \times \mathbf{b}| = 2\sin\delta\sqrt{1 + \cos^2\delta}$

1. Calculate the vector product $\mathbf{a}\times\mathbf{b}$ and its magnitude (deriving the result of part a).

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

Using factor formulae and $\theta-\phi = -2\delta$:
$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} \sin(-2\delta) \\ -2\sin\left(\frac{\theta+\phi}{2}\right)\sin\left(\frac{\theta-\phi}{2}\right) \\ 2\cos\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right) \end{pmatrix} = \begin{pmatrix} -2\sin\delta\cos\delta \\ 2\sin\left(\frac{\phi+\theta}{2}\right)\sin\delta \\ 2\cos\left(\frac{\phi+\theta}{2}\right)\sin\delta \end{pmatrix}$

$|\mathbf{a}\times\mathbf{b}|^2 = 4\sin^2\delta\cos^2\delta + 4\sin^2\delta \left( \sin^2\left(\frac{\phi+\theta}{2}\right) + \cos^2\left(\frac{\phi+\theta}{2}\right) \right)$
$|\mathbf{a}\times\mathbf{b}|^2 = 4\sin^2\delta(1+\cos^2\delta) \implies |\mathbf{a}\times\mathbf{b}| = 2\sin\delta\sqrt{1+\cos^2\delta}$

2. Find the magnitudes of the original 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. Use the geometric definition of the vector product magnitude to deduce $\sin\alpha$.

$|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\alpha$

Substituting the derived values:
$2\sin\delta\sqrt{1+\cos^2\delta} = (\sqrt{2})(\sqrt{2})\sin\alpha$
$2\sin\delta\sqrt{1+\cos^2\delta} = 2\sin\alpha$
$\sin\alpha = \sin\delta\sqrt{1+\cos^2\delta} \quad \text{(Shown)}$