6. (a) (i) By writing $3\theta = (2\theta + \phi)$, 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 6 - 2009 - Paper 2

To solve the equation $8 \sin^3 \theta - 6 \sin \theta + 1 = 0$, we start by substituting $x = \sin \theta$:

$8x^3 - 6x + 1 = 0.$

This is a cubic equation, which can be solved using various methods (e.g., numerical methods, factorization, or graphical). One possible root can be found using trial and error for rational solutions. Upon testing, we find:

Let $x = \frac{1}{2}$:

$8(\frac{1}{2})^3 - 6(\frac{1}{2}) + 1 = 8(\frac{1}{8}) - 3 + 1 = 1 - 3 + 1 = -1,$
which means $x = \frac{1}{2}$ is not a root. We try next $x = \frac{3}{4}$:

$8(\frac{3}{4})^3 - 6(\frac{3}{4}) + 1 = 8(\frac{27}{64}) - 6(\frac{3}{4}) + 1 = \frac{216}{64} - \frac{96}{64} + \frac{64}{64} = \frac{216 - 96 + 64}{64} = \frac{184}{64} = 0,$ thus, $\sin \theta = \frac{3}{4}$.

Hence, to find $\theta$, we evaluate: $\theta = \arcsin(\frac{3}{4}).$

To show that $\sin 15^{\circ} = \frac{1}{4}(\sqrt{6} - \sqrt{2})$, we can use the angle subtraction formula. First, recognize that $15^{\circ} = 45^{\circ} - 30^{\circ}$.

Using the formula: $\sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha,$ we set $\theta = 45^{\circ}$ and $\alpha = 30^{\circ}$. Thus, we have:

$\sin 15^{\circ} = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ}.$

Substituting the known trigonometric values: $\sin 45^{\circ} = \frac{\sqrt{2}}{2}, \quad \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \quad \cos 45^{\circ} = \frac{\sqrt{2}}{2}, \quad \sin 30^{\circ} = \frac{1}{2},$
this gives: $\sin 15^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}.$

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