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 $0 < x < 180^ ext{o}$, $$\frac{1 - ext{cos} 4x + ext{sin} 4x}{1 + ext{cos} 4x + ext{sin} 4x} = 3 \sin 2x$$ giving your answers to one decimal place where appropriate.

A-Level Edexcel Maths Pure Modelling with Functions: In this question you should show

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

Question 12

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 $1 + ext{cos} 2 heta + ext{sin} 2 heta eq 0$ 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 identity, start by using double angle formulas:

Recall the double angle formulas:
- $\text{cos} 2\theta = 2 \text{cos}^2 \theta - 1$
- $\text{sin} 2\theta = 2 \text{sin} \theta \text{cos} \theta$
Substitute these into the left-hand side:

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

Simplifying gives:

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

which can be factored:

$\frac{2(1 - \text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta)}{2(\text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta)} = \frac{1 - \text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta}{\text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta}$
Recognize $1 - \text{cos}^2 \theta = \text{sin}^2 \theta$ , thus:

$\frac{\text{sin}^2 \theta + \text{sin} \theta \text{cos} \theta}{\text{cos}^2 \theta + \text{sin} \theta \text{cos} \theta}$
Factor out $\text{sin} \theta$ in the numerator and denominator:

$\frac{\text{sin} \theta(\text{sin} \theta + \text{cos} \theta)}{\text{cos} \theta(\text{cos} \theta + \text{sin} \theta)}$
This reduces to:

$\tan \theta$
Hence, the identity is proven.

Step 2

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

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Answer

Using the result from part (a), we can rewrite:

$\frac{1 - \text{cos} 4x + \text{sin} 4x}{1 + \text{cos} 4x + \text{sin} 4x} = \tan 2x$

Now, we equate:

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

This can be solved by rewriting:

$\tan 2x = \frac{\sin 2x}{\cos 2x}$

Thus:

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

Rearranging gives:

$\sin 2x = 3 \sin 2x \cos 2x$

Now, simplifying:

$\sin 2x(1 - 3 \cos 2x) = 0$

This gives two cases to consider:

Case 1: $\sin 2x = 0$ .
- For $0 < 2x < 360^ ext{o}$ $0 < 2 x < 36 0^{e} x t o$ ,
  - Thus, $2x = 0^ ext{o}, 180^ ext{o}$ ,
  - Which gives $x = 0^ ext{o}, 90^ ext{o}$ . (only $90^ ext{o}$ is valid)
Case 2: $1 - 3 \cos 2x = 0$ .
- Thus, $\cos 2x = \frac{1}{3}$ .

Using: $\cos^{-1}(\frac{1}{3})$ , we find:

$2x \approx 70.53^ ext{o}$ (in the first quadrant) and $2x \approx 289.47^ ext{o}$ (which gives a corresponding $x$ of $144.735^ ext{o}$ ). Thus, valid solutions for $x$ are
$x \approx 35.27^ ext{o}, 72.36^ ext{o}$ .

Thus combining all valid values gives:

$x = 90^ ext{o}, 35.3^ ext{o}, 144.7^ ext{o}$ .

Final solutions rounded to one decimal place: $x \approx 90.0^ ext{o}, 35.3^ ext{o}, 144.7^ ext{o}$ .

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

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

Given that $1 + ext{cos} 2 heta + ext{sin} 2 heta eq 0$ 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 $0 < x < 180^ ext{o}$, $\frac{1 - \text{cos} 4x + \text{sin} 4x}{1 + \text{cos} 4x + \text{sin} 4x} = 3 \sin 2x$

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