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 5 - 2005 - Paper 5

Starting with the left side of the equation:

1. Recall that ( sin 2θ = 2 sin θ cos θ ), so

   $$2(2 sin θ cos θ) - 3 cos 2θ - 3 sin θ + 3 = 4 sin θ cos θ - 3(2 cos² θ - 1) - 3 sin θ + 3.$$

2. Expanding this gives:

   $$4 sin θ cos θ - 6 cos² θ + 3 - 3 sin θ + 3 = 4 sin θ cos θ - 6 cos² θ - 3 sin θ + 6.$$

3. To show the original equation:

   $$= sin θ(4 cos θ + 6 sin θ - 3)$$

This establishes the relationship.

To find R and α, we use the formula:

1. Set:

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

2. Find α:

   $$tan α = \frac{6}{4} = \frac{3}{2}$$

3. Thus, α = \arctan \left(\frac{3}{2}\right). Hence,

   $$4 cos θ + 6 sin θ = R sin(θ + α).$$

To solve this equation:

1. Start from:

   $$2 sin 2θ = \frac{3}{2}(cos 2θ + sin θ - 1)$$

2. Substitute sin 2θ:

   $$2(2 sin θ cos θ) = \frac{3}{2}(2 cos² θ - 1 + sin θ - 1)$$

3. Rearranging gives:

   $$4 sin θ cos θ = 3 cos² θ + 3 sin θ - 3.$$

4. Solve for θ over the interval [0, π]: Finding the exact solutions leads to:

   $$θ = 2.12, 0.588 ext{ (and other possible angles based on period 2π)}.$$