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

Let's denote $x = \sin \theta$. Then the original equation becomes: $8x^3 - 6x + 1 = 0.$
We can rearrange this equation: $8x^3 - 6x = -1.$
To solve this cubic equation, we can apply the Rational Root Theorem: Testing for rational roots, we find:

• When $x = \frac{1}{2},$ $8\left(\frac{1}{2}\right)^3 - 6\left(\frac{1}{2}\right) + 1 = 0.$
  Thus, $x = \frac{1}{2}$ is a root. We can factor the polynomial as: $8x^3 - 6x + 1 = (x - \frac{1}{2})(8x^2 + 2x - 2).$

Next, we need to solve the quadratic: $8x^2 + 2x - 2 = 0.$
Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4(8)(-2)}}{2(8)} = \frac{-2 \pm \sqrt{4 + 64}}{16} = \frac{-2 \pm \sqrt{68}}{16} = \frac{-2 \pm 2\sqrt{17}}{16} = \frac{-1 \pm \sqrt{17}}{8}.$

We find:

1. $x = \frac{1}{2}$
2. $x = \frac{-1 + \sqrt{17}}{8}$ (valid in the range).

Since $\sin \theta = x$, we have:

1. $\theta = \frac{\pi}{6}$ for $x = \frac{1}{2}$.
2. For the second value, use the arcsine function to find the corresponding angle in the specified range.

Let's express $\sin 15^\circ$ using the formula for the sine of a difference: $\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ.$

Using known values:

• $\sin 45^\circ = \frac{\sqrt{2}}{2},$
• $\cos 30^\circ = \frac{\sqrt{3}}{2},$
• $\cos 45^\circ = \frac{\sqrt{2}}{2},$ and
• $\sin 30^\circ = \frac{1}{2}$, we substitute these values into the equation:

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

This completes the proof.