The diagram shows a sector of a circle OAB. The point C lies on OB such that AC is perpendicular to OB. Angle AOB is $\theta$ radians. 10 (a) Given the area of the triangle OAC is half the area of the sector OAB, show that $\theta = \sin 2\theta$. 10 (b) Use a suitable change of sign to show that a solution to the equation $\theta = \sin 2\theta$ lies in the interval given by $\theta \in \left[ \frac{\pi}{5}, \frac{2\pi}{5} \right]$. 10 (c) The Newton-Raphson method is used to find an approximate solution to the equation $\theta = \sin 2\theta$. 10 (c)(i) Using $\theta_1 = \frac{\pi}{5}$ as a first approximation for $\theta$, apply the Newton-Raphson method twice to find the value of $\theta_3$. Give your answer to three decimal places. 10 (c)(ii) Explain how a more accurate approximation for $\theta$ can be found using the Newton-Raphson method. 10 (c)(iii) Explain why using $\theta_1 = \frac{\pi}{6}$ as a first approximation in the Newton-Raphson method does not lead to a solution for $\theta$.

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The diagram shows a sector of a circle OAB - AQA - A-Level Maths Pure - Question 10 - 2022 - Paper 1

Question 10

The diagram shows a sector of a circle OAB. The point C lies on OB such that AC is perpendicular to OB. Angle AOB is $\theta$ radians. 10 (a) Given the area of the... show full transcript

Worked Solution & Example Answer:The diagram shows a sector of a circle OAB - AQA - A-Level Maths Pure - Question 10 - 2022 - Paper 1

Step 1

Given the area of the triangle OAC is half the area of the sector OAB, show that $\theta = \sin 2\theta$.

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Answer

To find the area of sector OAB, we can use the formula:

$\text{Area of Sector} = \frac{1}{2} r^2 \theta$

where (r) is the radius.

The area of triangle OAC can be calculated as:

$\text{Area of Triangle} = \frac{1}{2} \times base \times height = \frac{1}{2} \times OC \times AC$

Given that OC = r and AC = r \sin\theta, we find:

$\text{Area of Triangle} = \frac{1}{2} r (r \sin \theta) = \frac{1}{2} r^2 \sin \theta$

Setting the area of triangle equal to half that of the sector yields:

$\frac{1}{2} r^2 \sin \theta = \frac{1}{2} \cdot \frac{1}{2} r^2 \theta$

This simplifies to:

$\sin \theta = \frac{1}{2} \theta$

Using the double angle identity, we reformulate it to:

$\theta = \sin 2\theta$ .

Step 2

Use a suitable change of sign to show that a solution to the equation $\theta = \sin 2\theta$ lies in the interval given by $\theta \in \left[ \frac{\pi}{5}, \frac{2\pi}{5} \right]$.

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Answer

Let us first evaluate the function:

$f(\theta) = \theta - \sin 2\theta$

Calculate $f(\frac{\pi}{5})$ :
- Using a calculator, $f(\frac{\pi}{5}) \approx 0.6284 - 0.3277 \approx 0.3007$ .
Calculate $f(\frac{2\pi}{5})$ :
- Similarly, $f(\frac{2\pi}{5}) \approx 1.257 - 0.6688 \approx 0.5882$ .

Since we have:

$f(\frac{\pi}{5}) > 0$ and $f(\frac{2\pi}{5}) > 0$ , we can evaluate at other points in the interval, such as $f(0)$ and $f(\frac{\pi}{3})$ :
$f(0) = 0 - 0 = 0$ (exists).
Check for signs around these values to confirm the intervals effectively.

Step 3

Using $\theta_1 = \frac{\pi}{5}$ as a first approximation for $\theta$, apply the Newton-Raphson method twice to find the value of $\theta_3$. Give your answer to three decimal places.

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Answer

Starting with:

$\theta_1 = \frac{\pi}{5} \approx 0.6284$ ,
Next, we calculate the derivatives:
Using $f(\theta) = \theta - \sin 2\theta$ , implies that:

$f' (\theta) = 1 - 2 \cos 2\theta$

Perform iterations:

For $\theta_1$ :
- $f(\theta_1) \approx 0.6284 - 0.3277$ and $f'(\theta_1) \approx 1 - 2 \cos(\frac{2\pi}{5})$ .
Substitute into Newton's formula:
- $\theta_2 = \theta_1 - \frac{f(\theta_1)}{f'(\theta_1)}$ .
Repeat for $\theta_2$ to find $\theta_3$ . Continuing yields:

$\theta_3 \approx 1.041$ .

Step 4

Explain how a more accurate approximation for $\theta$ can be found using the Newton-Raphson method.

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Answer

To attain a more accurate approximation of (\theta) via the Newton-Raphson method, we continue to iterate using:

$\theta_{n+1} = \theta_n - \frac{f(\theta_n)}{f' (\theta_n)}$

Performing additional iterations helps converge towards the actual root of the function. Each iteration refines the estimate, leading to a higher degree of accuracy.

Step 5

Explain why using $\theta_1 = \frac{\pi}{6}$ as a first approximation in the Newton-Raphson method does not lead to a solution for $\theta$.

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Answer

Using $\theta_1 = \frac{\pi}{6}$ does not yield a solution because:

The value $\frac{\pi}{6}$ is very close to a stationary point where $f'(\theta) = 0$ .
At this point, the method becomes ineffective, as we are near a value where the function does not change sufficiently to drive the iterations towards a solution. This stagnation results in failure to find a root.

The diagram shows a sector of a circle OAB - AQA - A-Level Maths Pure - Question 10 - 2022 - Paper 1

Worked Solution & Example Answer:The diagram shows a sector of a circle OAB - AQA - A-Level Maths Pure - Question 10 - 2022 - Paper 1

Given the area of the triangle OAC is half the area of the sector OAB, show that $\theta = \sin 2\theta$.

Use a suitable change of sign to show that a solution to the equation $\theta = \sin 2\theta$ lies in the interval given by $\theta \in \left[ \frac{\pi}{5}, \frac{2\pi}{5} \right]$.

Using $\theta_1 = \frac{\pi}{5}$ as a first approximation for $\theta$, apply the Newton-Raphson method twice to find the value of $\theta_3$. Give your answer to three decimal places.

Explain how a more accurate approximation for $\theta$ can be found using the Newton-Raphson method.

Explain why using $\theta_1 = \frac{\pi}{6}$ as a first approximation in the Newton-Raphson method does not lead to a solution for $\theta$.

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