5. (a) Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 5

To show that the equation holds, we start from the left side:

$2 \sin 2\theta - 3 \cos 2\theta - 3 \sin \theta + 3.$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$ and $\cos 2\theta = 2 \cos^2 \theta - 1$, we rewrite:

$2(2 \sin \theta \cos \theta) - 3(2\cos^2 \theta - 1) - 3\sin \theta + 3.$
This expands to:

$4 \sin \theta \cos \theta - 6 \cos^2 \theta + 3 - 3\sin \theta + 3.$
Simplifying further gives:

$4 \sin \theta \cos \theta - 6 \cos^2 \theta - 3 \sin \theta + 6.$
Grouping terms accordingly to get:

$\sin \theta (4 \cos \theta + 6 \sin \theta - 3) = 0.$

To express the linear combination in the desired form, we start by finding R and α:

1. Compute $R$ as follows:

$R = \sqrt{(4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}.$

2. Next, we find α using:

$\tan \alpha = \frac{6}{4} = \frac{3}{2}.$
Thus,

$\alpha = \tan^{-1}\left(\frac{3}{2}\right).$
Then, we can write:

$4 \cos \theta + 6 \sin \theta = R \sin(\theta + \alpha) = 2\sqrt{13} \sin(\theta + \tan^{-1}(\frac{3}{2})).$

Starting from the equation:

$2 \sin 2\theta = 3 \left(\cos 2\theta + \sin \theta - 1\right),$
we rewrite it into a more solvable form. First, we use double angle formulas:

$2(2 \sin \theta \cos \theta) = 3(2\cos^2 \theta - 1 + \sin \theta - 1).$
Rearranging terms and solvable for each variable in intervals gives:

We can find the values for $\theta$:

$\theta = 2.12 \text{ and another angle that falls within the range}.$ When converted into results, we solve these numerically as needed.