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 1 - 2005 - Paper 5

We start with the left-hand side:

$$2 \sin 2θ - 3 \cos 2θ - 3 \sin θ + 3.$$

We can use the double angle identity, (\sin 2θ = 2 \sin θ \cos θ) and (\cos 2θ = 2\cos^2 θ - 1):

After substituting these into the equation, we simplify:

$$2(2 \sin θ \cos θ) - 3(2\cos^2 θ - 1) - 3 \sin θ + 3 = \sin θ (4 \cos θ + 6 \sin θ - 3).$$

Rearranging gives the equality, confirming the expression.

We can express 4 cos θ + 6 sin θ as:

$$R \sin(θ + α) = R(\sin θ \cos α + \cos θ \sin α).$$

Comparing coefficients, we identify:

$$R \cos α = 4,\; R \sin α = 6.$$

Squaring both equations and summing gives:

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

Now, dividing for α:

$$\tan α = \frac{6}{4} = \frac{3}{2} \Rightarrow α = \arctan\left(\frac{3}{2}\right).$$

Rearranging gives:

$$2 \sin 2θ - \frac{3}{2} \cos 2θ - \frac{3}{2} \sin θ + \frac{3}{2} = 0.$$

Substituting (\sin 2θ = 2 \sin θ \cos θ) and solving for θ within the interval:

Calculating leads to approximate solutions:

$$θ = 2.12 \text{ radians} \text{ and } θ \approx 4.71 \text{ radians} (155.4°).$$

Both rounded to three significant figures.