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

Starting with the left-hand side:

$$2 sin 2θ - 3 cos 2θ - 3 sin θ + 3,$$

we know that:

$$sin 2θ = 2 sin θ cos θ,$$

which allows us to rewrite it as:

$$2(2 sin θ cos θ) - 3(2 cos^2 θ - 1) - 3 sin θ + 3.$$

This simplifies to:

$$4 sin θ cos θ - 6 cos^2 θ + 3 - 3 sin θ.$$

Grouping terms, we have:

$$4 sin θ cos θ - 3 sin θ + 3 - 6 cos^2 θ.$$

Hence, factor out sin θ from the end:

$$sin θ(4 cos θ + 6 sin θ - 3).$$

To express this in the desired form, we first need to calculate R:

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

Next, we determine α using:

$$tan(α) = \frac{b}{a} = \frac{6}{4} = \frac{3}{2}.$$

Calculating α gives:

$$α = arctan(\frac{3}{2}) ≈ 0.588 ext{ radians} (allow 33.79°).$$

Start by rearranging the equation:

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

which simplifies to:

$$sin(2θ) = \frac{3}{4}(cos 2θ + sin θ - 1).$$

Use the identity sin(2θ) = 2 sin θ cos θ on the left:

$$2 sin θ cos θ = \frac{3}{4}(cos(2θ) + sin(θ) - 1).$$

By solving this numerically, we analyze:

• The solutions yield approximations in radians:
  • θ ≈ 2.12.
• Up to 3 significant figures gives us:
  • θ ≈ 0.588.
  • θ ≈ 155.4°.