A-Level Edexcel Maths: Pure Trigonometric Proof: 5. (a) Using the identity cos(A

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

Question 5

5. (a) Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A. (b) Show that 2 sin 2θ - 3 cos 2θ - 3 sin θ + 3 = sin θ... show full transcript

Worked Solution & Example Answer: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

Step 1

Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A.

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Answer

To prove that, we can start from the given identity:

cos 2A = cos(A + A) = cos A cos A - sin A sin A = cos² A - sin² A = (1 - sin² A) - sin² A = 1 - 2sin² A.

This completes the proof.

Step 2

Show that 2 sin 2θ - 3 cos 2θ - 3 sin θ + 3 = sin θ(4 cos θ + 6 sin θ - 3).

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Answer

Starting with the left side of the equation:

Recall that ( sin 2θ = 2 sin θ cos θ ), so
$2(2 sin θ cos θ) - 3 cos 2θ - 3 sin θ + 3 = 4 sin θ cos θ - 3(2 cos² θ - 1) - 3 sin θ + 3.$
Expanding this gives:
$4 sin θ cos θ - 6 cos² θ + 3 - 3 sin θ + 3 = 4 sin θ cos θ - 6 cos² θ - 3 sin θ + 6.$
To show the original equation:
$= sin θ(4 cos θ + 6 sin θ - 3)$

This establishes the relationship.

Step 3

Express 4 cos θ + 6 sin θ in the form R sin(θ + α), where R > 0 and 0 < α < \frac{\pi}{2}.

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Answer

To find R and α, we use the formula:

Set:
$R = \sqrt{(4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$
Find α:
$tan α = \frac{6}{4} = \frac{3}{2}$
Thus, α = \arctan \left(\frac{3}{2}\right). Hence,
$4 cos θ + 6 sin θ = R sin(θ + α).$

Step 4

Hence, for 0 ≤ θ < π, solve 2 sin 2θ = \frac{3}{2}(cos 2θ + sin θ - 1).

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Answer

To solve this equation:

Start from:
$2 sin 2θ = \frac{3}{2}(cos 2θ + sin θ - 1)$
Substitute sin 2θ:
$2(2 sin θ cos θ) = \frac{3}{2}(2 cos² θ - 1 + sin θ - 1)$
Rearranging gives:
$4 sin θ cos θ = 3 cos² θ + 3 sin θ - 3.$
Solve for θ over the interval [0, π]: Finding the exact solutions leads to:
$θ = 2.12, 0.588 ext{ (and other possible angles based on period 2π)}.$

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

Worked Solution & Example Answer: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

Using the identity cos(A + B) = cos A cos B - sin A sin B, prove that cos 2A = 1 - 2 sin² A.

Show that 2 sin 2θ - 3 cos 2θ - 3 sin θ + 3 = sin θ(4 cos θ + 6 sin θ - 3).

Express 4 cos θ + 6 sin θ in the form R sin(θ + α), where R > 0 and 0 < α < \frac{\pi}{2}.

Hence, for 0 ≤ θ < π, solve 2 sin 2θ = \frac{3}{2}(cos 2θ + sin θ - 1).

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