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 θ radians. 10 (a) Given the area of the triangle OAC is half the area of the sector OAB, show that θ = sin 2θ 10 (b) Use a suitable change of sign to show that a solution to the equation θ = sin 2θ lies in the interval given by θ ∈ [ rac{ heta}{5}, rac{2 heta}{5} ] 10 (c) The Newton-Raphson method is used to find an approximate solution to the equation θ = sin 2θ 10 (c)(i) Using θ₁ = rac{ heta}{5} as a first approximation for θ, apply the Newton-Raphson method twice to find the value of θ₃. Give your answer to three decimal places. 10 (c)(ii) Explain how a more accurate approximation for θ can be found using the Newton-Raphson method. 10 (c)(iii) Explain why using θ₁ = rac{ heta}{6} as a first approximation in θ does not lead to a solution for θ.

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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 θ radians. 10 (a) Given the area of the triangl... 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 θ = sin 2θ

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Answer

The area of sector OAB is given by: $\text{Area} = \frac{1}{2} r^2 \theta$ For triangle OAC, we can express its area as: $\text{Area} = \frac{1}{2} \cdot OC \cdot AC$ Since AC is perpendicular to OB, the area can also be expressed as: $\text{Area} = \frac{1}{2} r \cdot r \sin(θ) = \frac{1}{2} r^2 \sin(θ)$ Setting the area of the triangle equal to half the area of the sector, we have: $\frac{1}{2} r^2 \sin(θ) = \frac{1}{4} r^2 \theta$ By simplifying, we can cancel out ( \frac{1}{2} r^2 ) from both sides, leading to: $\sin(θ) = \frac{1}{2} \theta$ Using the double angle formula, we can express this as: $\theta = \sin(2θ)$ .

Step 2

Use a suitable change of sign to show that a solution to the equation θ = sin 2θ lies in the interval given by θ ∈ [ rac{π}{5}, rac{2 heta}{5} ]

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Answer

To find the change of sign, we evaluate the function: $f(θ) = θ - \sin(2θ)$ Calculate:\

At ( θ = \frac{π}{5} ): $f(\frac{π}{5}) = \frac{π}{5} - \sin(\frac{2π}{5})$ , where this value is negative.
At ( θ = \frac{2 heta}{5} ): $f(\frac{2π}{5}) = \frac{2π}{5} - \sin(\frac{4π}{5})$ , where this value is positive. Since ( f(\frac{π}{5}) < 0 ) and ( f(\frac{2π}{5}) > 0 ), there exists at least one root in the interval [ \frac{π}{5}, \frac{2 heta}{5} ] by the Intermediate Value Theorem.

Step 3

Using θ₁ = π/5 as a first approximation for θ, apply the Newton-Raphson method twice to find the value of θ₃.

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Answer

The Newton-Raphson formula is given by: $θ_{n+1} = θ_n - \frac{f(θ_n)}{f'(θ_n)}$ First, compute:

At ( θ_1 = \frac{π}{5} ):
- Find ( f(θ_1) ) and ( f'(θ_1) ).
Use these values to find ( θ_2 ). Then, for ( θ_2 ), repeat the evaluation to find ( θ_3 ). Final Result:
Value of ( θ_3 ) should be given to three decimal places.

Step 4

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

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Answer

A more accurate approximation can be found by continuing the iterations of the Newton-Raphson method. By consistently applying the formula and re-evaluating for each subsequent approximation, the solution will converge to a more precise root of the equation.

Step 5

Explain why using θ₁ = π/6 as a first approximation in θ does not lead to a solution for θ.

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Answer

Using θ₁ = π/6 does not lead to a solution because:

At this value, the derivative ( f'(θ) ) evaluates to zero, indicating a stationary point.
The function may not cross the x-axis at this point, hence no change of sign occurs around this approximation, leading to failure in convergence.

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 θ = sin 2θ

Use a suitable change of sign to show that a solution to the equation θ = sin 2θ lies in the interval given by θ ∈ [ rac{π}{5}, rac{2 heta}{5} ]

Using θ₁ = π/5 as a first approximation for θ, apply the Newton-Raphson method twice to find the value of θ₃.

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

Explain why using θ₁ = π/6 as a first approximation in θ does not lead to a solution for θ.

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