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In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 12 - 2021 - Paper 1

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In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. (a) Given that $1 + ext{cos}2 ... show full transcript

Worked Solution & Example Answer:In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 12 - 2021 - Paper 1

Step 1

Given that $1 + \text{cos}2\theta + \text{sin}2\theta \neq 0$ prove that $$\frac{1 - \text{cos}2\theta + \text{sin}2\theta}{1 + \text{cos}2\theta + \text{sin}2\theta} = \tan \theta$$

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Answer

To prove the equation, we start by applying the double angle formulae:

  1. Recall that:

    • cos2θ=2cos2θ−1\text{cos}2\theta = 2\text{cos}^2\theta - 1, and
    • sin2θ=2sinθcosθ\text{sin}2\theta = 2\text{sin}\theta \text{cos}\theta.

  2. Substitute these into the original expression: 1−(2cos2θ−1)+2sinθcosθ1+(2cos2θ−1)+2sinθcosθ\frac{1 - (2\text{cos}^2\theta - 1) + 2\text{sin}\theta \text{cos}\theta}{1 + (2\text{cos}^2\theta - 1) + 2\text{sin}\theta \text{cos}\theta}

  3. Simplifying the numerator:

    • 1+1−2cos2θ+2sinθcosθ=2−2cos2θ+2sinθcosθ1 + 1 - 2\text{cos}^2\theta + 2\text{sin}\theta \text{cos}\theta = 2 - 2\text{cos}^2\theta + 2\text{sin}\theta \text{cos}\theta.

  4. Simplifying the denominator:

    • 1−1+2cos2θ+2sinθcosθ=2cos2θ+2sinθcosθ1 - 1 + 2\text{cos}^2\theta + 2\text{sin}\theta \text{cos}\theta = 2\text{cos}^2\theta + 2\text{sin}\theta \text{cos}\theta.

  5. Now the fraction becomes: 2−2cos2θ+2sinθcosθ2cos2θ+2sinθcosθ=2(1−cos2θ+sinθcosθ)2(cos2θ+sinθcosθ)=1−cos2θ+sinθcosθcos2θ+sinθcosθ.\frac{2 - 2\text{cos}^2\theta + 2\text{sin}\theta \text{cos}\theta}{2\text{cos}^2\theta + 2\text{sin}\theta \text{cos}\theta} = \frac{2(1 - \text{cos}^2\theta + \text{sin}\theta \text{cos}\theta)}{2(\text{cos}^2\theta + \text{sin}\theta \text{cos}\theta)} = \frac{1 - \text{cos}^2\theta + \text{sin}\theta \text{cos}\theta}{\text{cos}^2\theta + \text{sin}\theta \text{cos}\theta}.

  6. Recognizing that 1−cos2θ=sin2θ1 - \text{cos}^2\theta = \text{sin}^2\theta, we get: sin2θ+sinθcosθcos2θ+sinθcosθ=tan⁡θ.\frac{\text{sin}^2\theta + \text{sin}\theta \text{cos}\theta}{\text{cos}^2\theta + \text{sin}\theta \text{cos}\theta} = \tan \theta.

Thus, we have proven the equation as required.

Step 2

Hence solve, for $0 < x < 180^{\circ}$ $$\frac{1 - \text{cos}4x + \text{sin}4x}{1 + \text{cos}4x + \text{sin}4x} = 3 \sin 2x$$

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Answer

Using the result from part (a), we substitute tan⁡\tan into the formula:

  1. From part (a): 1−cos4x+sin4x1+cos4x+sin4x=tan⁡(2x).\frac{1 - \text{cos}4x + \text{sin}4x}{1 + \text{cos}4x + \text{sin}4x} = \tan(2x).

  2. Therefore, we have: tan⁡(2x)=3sin⁡(2x).\tan(2x) = 3\sin(2x).

  3. To solve for xx, recall that: tan⁡(2x)=sin⁡(2x)cos⁡(2x).\tan(2x) = \frac{\sin(2x)}{\cos(2x)}. Then, equating: sin⁡(2x)cos⁡(2x)=3sin⁡(2x)  ⟹  1=3cos⁡(2x).\frac{\sin(2x)}{\cos(2x)} = 3\sin(2x) \implies 1 = 3\cos(2x).

  4. This rearranges to: cos⁡(2x)=13.\cos(2x) = \frac{1}{3}.

  5. Now, we can find 2x2x: 2x=cos⁡−1(13) or 360∘−cos⁡−1(13).2x = \cos^{-1}\left(\frac{1}{3}\right) \text{ or } 360^{\circ} - \cos^{-1}\left(\frac{1}{3}\right).

  6. Thus, we find: 2x≈70.5288∘extand2x≈289.4712∘.2x \approx 70.5288^{\circ} ext{ and } 2x \approx 289.4712^{\circ}. Therefore, dividing by 2 gives: x≈35.3∘extandx≈144.7∘.x \approx 35.3^{\circ} ext{ and } x \approx 144.7^{\circ}.

Finally, the solutions are:

  • x≈35.3∘x \approx 35.3^{\circ}
  • x≈144.7∘x \approx 144.7^{\circ}.

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