Show that $\cos 3\theta = 4\cos^{3}\theta - 3\cos\theta$. Hence, or otherwise, evaluate $\displaystyle \int_{0}^{\pi/6} \sin 2\theta\,\cos 3\theta\, d\theta$ exactly.

1. Express $\cos 3\theta$ in terms of $2\theta$ and $\theta$, then apply the compound angle formula.

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

2. Apply the double angle formulas to express the terms using single angles.

$= (2\cos^2\theta - 1)\cos \theta - (2\sin \theta \cos \theta)\sin \theta$

3. Expand the expression and use the Pythagorean identity $\sin^2\theta = 1 - \cos^2\theta$ to simplify entirely in terms of $\cos\theta$.

$= 2\cos^3\theta - \cos \theta - 2\sin^2\theta \cos \theta$
$= 2\cos^3\theta - \cos \theta - 2(1 - \cos^2\theta)\cos \theta$
$= 2\cos^3\theta - \cos \theta - 2\cos \theta + 2\cos^3\theta$
$= 4\cos^3\theta - 3\cos \theta$

1. Express the integrand in terms of $\sin\theta$ and $\cos\theta$ using the identity from part (a) and the double angle formula.

Using $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$:
$\sin 2\theta \cos 3\theta = (2\sin\theta\cos\theta)(4\cos^3\theta - 3\cos\theta)$

2. Expand the expression to set up for integration.

$(2\sin\theta\cos\theta)(4\cos^3\theta - 3\cos\theta) = 8\cos^4\theta\sin\theta - 6\cos^2\theta\sin\theta$

3. Integrate the expanded expression by recognizing the standard form $\int [f(x)]^n f'(x) \, dx$.

$\int_{0}^{\pi/6} (8\cos^4\theta\sin\theta - 6\cos^2\theta\sin\theta) \, d\theta = \left[ -\frac{8}{5}\cos^5\theta + 2\cos^3\theta \right]_{0}^{\pi/6}$

4. Evaluate the definite integral using the limits exactly.

Substitute $\theta = \frac{\pi}{6}$ and $\theta = 0$:
$\left( -\frac{8}{5}\left(\frac{\sqrt{3}}{2}\right)^5 + 2\left(\frac{\sqrt{3}}{2}\right)^3 \right) - \left( -\frac{8}{5}(1)^5 + 2(1)^3 \right)$
$= \left( -\frac{8}{5}\left(\frac{9\sqrt{3}}{32}\right) + 2\left(\frac{3\sqrt{3}}{8}\right) \right) - \left( -\frac{8}{5} + 2 \right)$
$= \left( -\frac{9\sqrt{3}}{20} + \frac{15\sqrt{3}}{20} \right) - \frac{2}{5}$
$= \frac{6\sqrt{3}}{20} - \frac{2}{5}$
$= \frac{3\sqrt{3}}{10} - \frac{2}{5}$