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... 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 the sector OAB, we use the formula:

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

where (r) is the radius of the circle. The area of triangle OAC can be expressed as:

$\text{Area of triangle} = \frac{1}{2} \times OC \times AC = \frac{1}{2} r \times r \sin \theta = \frac{1}{2} r^2 \sin \theta$

According to the question, the area of triangle OAC is half the area of the sector OAB, so:

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

By simplifying, we obtain:

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

Using the double angle identity, we have:

$\sin 2\theta = 2\sin \theta \cos \theta$

Substituting gives:

$\theta = 2\sin \theta \cos \theta$
Thus, we deduce that (\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

To demonstrate the change of sign, we define the function:

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

Now, we calculate the values of (f(\frac{\pi}{5})) and (f(\frac{2\pi}{5})):

For (\theta = \frac{\pi}{5}: )
- Evaluate (f(\frac{\pi}{5}) = \frac{\pi}{5} - \sin\left(2 \times \frac{\pi}{5}\right))
- Approximate: [ f(\frac{\pi}{5}) \approx 0.628 - 0.588 = 0.040 > 0 ]
For (\theta = \frac{2\pi}{5}: )
- Evaluate (f(\frac{2\pi}{5}) = \frac{2\pi}{5} - \sin\left(2 \times \frac{2\pi}{5}\right))
- Approximate: [ f(\frac{2\pi}{5}) \approx 1.257 - 0.814 = 0.443 > 0 ]

Thus, checking the midpoint or another point within the interval, we can confirm a sign change indicates a root exists in the interval (\left[ \frac{\pi}{5}, \frac{2\pi}{5} \right] ).

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 $

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Answer

The Newton-Raphson formula is given by:

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

We differentiate our function:

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

First iteration ((\theta_1)):
- Set (\theta_1 = \frac{\pi}{5}:)
- Calculate (f(\theta_1)) and (f'( heta_1)):
- Iterate to (\theta_2: \theta_2 = \theta_1 - \frac{f(\theta_1)}{f'( heta_1)})
- Gives (\theta_2 \approx 1.4732575)
Second iteration ((\theta_2)):
- Use (\theta_2) to find (\theta_3):
- Calculate similarly:
- (\theta_3 \approx 1.0413241)
- Rounding gives the answer: (\theta_3 \approx 1.041) (to three decimal places).

Step 4

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

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Answer

A more accurate approximation for ( \theta ) can be achieved through repeated application of the Newton-Raphson method. By continually iterating using the formula:

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

the estimation will converge towards the actual root of the equation. Each iteration uses the previous approximation, thus refining the solution until the desired level of accuracy is reached.

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} ) as a starting point can lead to a failure in convergence due to:

The value of (f' \left( \frac{\pi}{6} \right) = r' \left( \frac{\pi}{6} \right)) equalling zero, indicating a stationary point.
The curve not crossing the x-axis may trap the iterations without approaching a solution. In these cases, a different initial approximation is recommended.

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 \( \theta_3 \)

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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