证明 $\frac{(\sin\theta-\cos\theta)(1+\sin\theta\cos\theta)}{\cos^{3}\theta}=\tan^{3}\theta-1.$ 由此，对于 $0<\theta<2\pi$，求满足 $(\sin\theta-\cos\theta)(1+\sin\theta\cos\theta)=\cos^{3}\theta$ 的优角 $\theta$。 求方程 $\sin3A\cos2A=\cos3A\sin2A$ 在 $0 \le A \le 180^\circ$ 范围内的解。 已知 $\sin x = -\tfrac{4}{5}$ 且 $\tan x > 0$，在不使用计算器的情况下，求 $\cos4x$ 的值。 在不使用计算器的情况下，证明 $\tan\frac{\pi}{12}=2-\sqrt{3}.$

1. Expand the numerator of the left-hand side

$\text{LHS} = \frac{\sin\theta + \sin^2\theta\cos\theta - \cos\theta - \sin\theta\cos^2\theta}{\cos^3\theta}$

2. Split the fraction into individual terms

$\text{LHS} = \frac{\sin\theta}{\cos^3\theta} + \frac{\sin^2\theta\cos\theta}{\cos^3\theta} - \frac{\cos\theta}{\cos^3\theta} - \frac{\sin\theta\cos^2\theta}{\cos^3\theta}$

3. Simplify each term using $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sec^2\theta = \frac{1}{\cos^2\theta}$

$\text{LHS} = \tan\theta\sec^2\theta + \tan^2\theta - \sec^2\theta - \tan\theta$

4. Apply the identity $\sec^2\theta = 1 + \tan^2\theta$ and simplify to reach the RHS

$\text{LHS} = \tan\theta(1+\tan^2\theta) + \tan^2\theta - (1+\tan^2\theta) - \tan\theta$
$= \tan\theta + \tan^3\theta + \tan^2\theta - 1 - \tan^2\theta - \tan\theta$
$= \tan^3\theta - 1 = \text{RHS}$

1. Use the result from part (a)(i) to simplify the equation

From (a)(i), we know that $(\sin\theta-\cos\theta)(1+\sin\theta\cos\theta) = \cos^3\theta(\tan^3\theta-1)$.
Substituting this into the given equation:
$\cos^3\theta(\tan^3\theta-1) = \cos^3\theta$
Assuming $\cos\theta \neq 0$ (since if $\cos\theta = 0$, $\sin\theta = \pm 1$, and LHS $\neq$ RHS), we can divide both sides by $\cos^3\theta$:
$\tan^3\theta - 1 = 1$
$\tan^3\theta = 2$

2. Find the basic angle for $\theta$

$\tan\theta = \sqrt[3]{2}$
Since $\tan\theta > 0$, $\theta$ lies in the 1st or 3rd quadrant.
The basic angle $\alpha$ is:
$\alpha = \tan^{-1}(\sqrt[3]{2}) \approx 0.89990 \text{ rad}$

3. Determine the reflex angle $\theta$ in the given domain

The question requires a reflex angle ($180^\circ < \theta < 360^\circ$ or $\pi < \theta < 2\pi$).
Therefore, $\theta$ must be in the 3rd quadrant.
$\theta = \pi + 0.89990 \approx 4.0415$
$\theta \approx 4.04 \text{ rad (to 3 s.f.)}$

1. Rearrange the equation and apply the compound angle formula for sine

$\sin 3A \cos 2A - \cos 3A \sin 2A = 0$
$\sin(3A - 2A) = 0$
$\sin A = 0$

2. Solve for $A$ in the given interval

For $0 \le A \le 180^\circ$,
$A = 0^\circ, 180^\circ$

1. Find the value of $\cos 2x$ using the double angle formula

$\cos 2x = 1 - 2\sin^2 x$
$\cos 2x = 1 - 2\left(-\frac{4}{5}\right)^2$
$\cos 2x = 1 - 2\left(\frac{16}{25}\right) = 1 - \frac{32}{25} = -\frac{7}{25}$

2. Find the value of $\cos 4x$ using the double angle formula again

$\cos 4x = 2\cos^2 2x - 1$
$\cos 4x = 2\left(-\frac{7}{25}\right)^2 - 1$
$\cos 4x = 2\left(\frac{49}{625}\right) - 1 = \frac{98}{625} - \frac{625}{625} = -\frac{527}{625}$

1. Express the angle as a difference of special angles

$\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$

2. Apply the compound angle formula for tangent

$\tan\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = \frac{\tan\frac{\pi}{4} - \tan\frac{\pi}{6}}{1 + \tan\frac{\pi}{4}\tan\frac{\pi}{6}}$

3. Substitute the exact values of the trigonometric functions

$\tan\frac{\pi}{4} = 1, \quad \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$

$\tan\frac{\pi}{12} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

4. Rationalise the denominator to show the final result

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2}$
$= \frac{3 - 2\sqrt{3} + 1}{3 - 1}$
$= \frac{4 - 2\sqrt{3}}{2}$
$= 2 - \sqrt{3}$