10. In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. (a) Given that $1 + ext{cos} 2 heta + ext{sin} 2 heta eq 0$ prove that $$\frac{1 - ext{cos} 2 heta + ext{sin} 2 heta}{1 + ext{cos} 2 heta + ext{sin} 2 heta} = \tan \theta$$ (b) Hence solve, for $0 < x < 180^{\circ}$, $$\frac{1 - ext{cos} 4x + ext{sin} 4x}{1 + ext{cos} 4x + ext{sin} 4x} = 3 \sin 2x$$ giving your answers to one decimal place where appropriate.

A-Level Edexcel Maths Pure Integration: 10. In this question you should

10. In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 11 - 2021 - Paper 1

Question 11

10. 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{co... show full transcript

Worked Solution & Example Answer:10. In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 11 - 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 this equation, we will use the double angle formulas for sine and cosine:

$\text{sin} 2\theta = 2 \text{sin} \theta \text{cos} \theta$
$\text{cos} 2\theta = 2\text{cos}^2 \theta - 1$

Substituting these into the left-hand side:

$\frac{1 - (2\text{cos}^2 \theta - 1) + 2\text{sin} \theta \text{cos} \theta}{1 + (2\text{cos}^2 \theta - 1) + 2\text{sin} \theta \text{cos} \theta}$

This simplifies to:

$\frac{2 - 2\text{cos}^2 \theta + 2\text{sin} \theta \text{cos} \theta}{2\text{cos}^2 \theta + 2\text{sin} \theta \text{cos} \theta}$

Factor out the common terms:

$\frac{2(1 - \text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta)}{2(\text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta)}$

Finally, recognizing that $1 - \text{cos}^2 \theta = \text{sin}^2 \theta$ , we have:

$\frac{\text{sin}^2 \theta + \text{sin} \theta \text{cos} \theta}{\text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta} = \tan \theta$ .

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

Utilizing the proven result from part (a), we can replace the left-hand side similarly:

$\frac{1 - \text{cos} 4x + \text{sin} 4x}{1 + \text{cos} 4x + \text{sin} 4x} = \tan(2x)$
This gives us:

$\tan(2x) = 3 \sin(2x)$

Now, we can rewrite the equation using the identity for tangent:

$\frac{\sin(2x)}{\cos(2x)} = 3 \sin(2x)$

Assuming $\sin(2x) \neq 0$ , we can divide both sides by $\sin(2x)$ , giving:

$\frac{1}{\cos(2x)} = 3$

This implies:

$\cos(2x) = \frac{1}{3}$

To find $2x$ , we take the inverse cosine:

$2x = \cos^{-1}(\frac{1}{3})$

This yields two solutions within the restriction $0 < 2x < 360^{\circ}$ :

$2x = \cos^{-1}(\frac{1}{3})$
$2x = 360^{\circ} - \cos^{-1}(\frac{1}{3})$

Consequently, solving for $x$ gives:

$x = \frac{\cos^{-1}(\frac{1}{3})}{2}$
$x = 180^{\circ} - \frac{\cos^{-1}(\frac{1}{3})}{2}$

Calculating these, we find:

$x \approx 35.3^{\circ}$
$x \approx 144.7^{\circ}$

Thus, to one decimal place, the answers are:

$x \approx 35.3^{\circ}$
$x \approx 144.7^{\circ}$ .

10. In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 11 - 2021 - Paper 1

Worked Solution & Example Answer:10. In this question you should show all stages of your working - Edexcel - A-Level Maths Pure - Question 11 - 2021 - Paper 1

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

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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