6. (a) (i) By writing $3\theta = (2\theta + \theta)$, show that $$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta.$$ (ii) Hence, or otherwise, for $0 < \theta < \frac{\pi}{3}$, solve $$8 \sin^3 \theta - 6 \sin \theta + 1 = 0.$$ Give your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 8 - 2009 - Paper 2

Let $x = \sin \theta$, we rewrite the equation as:

$8x^3 - 6x + 1 = 0.$
We can use the factor theorem to check for rational roots. Testing $x = \frac{1}{2}$:

$$8\left(\frac{1}{2}\right)^3 - 6\left(\frac{1}{2}\right) + 1 = 1 - 3 + 1 = -1 = 0.$So,$x - \frac{1}{2}$ is a factor. Dividing:

$8x^3 - 6x + 1 = (x - \frac{1}{2})(8x^2 + 4x - 2)$
We can solve the quadratic:

$8x^2 + 4x - 2 = 0.$
Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 64}}{16} = \frac{-4 \pm 8}{16}.$
This gives $x = \frac{1}{2}$ or $x = -\frac{3}{4}$ (not in range). Thus:

$\sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}.$
The only solution in the desired range is:

$\theta = \frac{\pi}{6}.$

To find $\sin 15^{\circ}$, we can express it as:

$\sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ}.$
Substituting known values:

$\sin 45^{\circ} = \frac{\sqrt{2}}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \cos 45^{\circ} = \frac{\sqrt{2}}{2}, \sin 30^{\circ} = \frac{1}{2}.$ Thus:

$\sin 15^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{1}{4}(\sqrt{6} - \sqrt{2}).$