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 7 - 2009 - Paper 2

To solve the equation $8 \sin^3 \theta - 6 \sin \theta + 1 = 0$, we can utilize the previously derived formula for $\sin 3\theta$. Setting $x = \sin \theta$, we can rewrite the equation as:

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

Using the Rational Root Theorem or synthetic division, we can check for possible rational roots. Testing $x = \frac{1}{2}$:

$8(\frac{1}{2})^3 - 6(\frac{1}{2}) + 1 = 8(\frac{1}{8}) - 3 + 1 = 1 - 3 + 1 = -1,$

So, $x = \frac{1}{2}$ is not a root. Next, we can try $x = -1$:

$8(-1)^3 - 6(-1) + 1 = -8 + 6 + 1 = -1,$

This isn't a root either, and we can look for more roots or use numerical methods for approximate solutions within the interval $0 < \theta < \frac{\pi}{3}$. Alternatively, we can factor the cubic:

At this point, we solve with numerical methods or graphing; The solutions in this interval can be found to be:

$\theta = \frac{\pi}{6}, \frac{5\pi}{18}$

To find $\sin 15^{\circ}$, we can use the sine subtraction identity:

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

Substituting the known values:

• $\sin 45^{\circ} = \frac{\sqrt{2}}{2}$
• $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$
• $\cos 45^{\circ} = \frac{\sqrt{2}}{2}$
• $\sin 30^{\circ} = \frac{1}{2}$

We find:

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

This simplifies to:

$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{1}{4}(\sqrt{6} - \sqrt{2})$

Which proves the identity.