15 (a) Show that \[ sin x - sin x \cos 2x \approx 2x^3 \] for small values of $x$. 15 (b) Hence, show that the area between the graph with equation \[ y = \sqrt{8(sin x - sin x \cos 2x)} \] the positive x-axis and the line $x = 0.25$ can be approximated by Area = $2^m \times 5^n$ where $m$ and $n$ are integers to be found. 15 (c) (i) Explain why \[ \int_{6.3}^{6.4} 2x^3 \, dx \] is not a suitable approximation for \[ \int_{6.3}^{6.4} (sin x - sin x \cos 2x) \, dx \] 15 (c) (ii) Explain how \[ \int_{6.3}^{6.4} (sin x - sin x \cos 2x) \, dx \] may be approximated by \[ \int_{a}^{b} 2x^3 \, dx \] for suitable values of $a$ and $b$.

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15 (a) Show that \[ sin x - sin x \cos 2x \approx 2x^3 \] for small values of $x$ - AQA - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Question 15

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15 (a) Show that \[ sin x - sin x \cos 2x \approx 2x^3 \] for small values of $x$. 15 (b) Hence, show that the area between the graph with equation \[ y = \sqr... show full transcript

Worked Solution & Example Answer:15 (a) Show that \[ sin x - sin x \cos 2x \approx 2x^3 \] for small values of $x$ - AQA - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Step 1

Show that $sin x - sin x \cos 2x \approx 2x^3$ for small values of $x$

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Answer

Using the small angle approximation for $sin x$ and the identity for $cos 2x$ , we have:

[ sin x \approx x \quad \text{and} \quad \cos 2x \approx 1 - (2x)^2/2 = 1 - 2x^2 ]

Thus,

[ sin x - sin x \cos 2x \approx x - x(1 - 2x^2) = x(2x^2) = 2x^3 ]

Step 2

Hence, show that the area can be approximated by Area = $2^m \times 5^n$

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Answer

First, we calculate the area under the curve from $x=0$ to $x=0.25$ :

[ A = \int_0^{0.25} \sqrt{8(2x^3)} , dx = \int_0^{0.25} 4\sqrt{2} x^{3/2} , dx = 4\sqrt{2} \cdot \frac{2}{5} x^{5/2} \biggr|_0^{0.25} ]

Evaluating this:

[ = 4\sqrt{2} \cdot \frac{2}{5} \cdot (0.25)^{5/2} = 4\sqrt{2} \cdot \frac{2}{5} \cdot \frac{1}{32} = \frac{8\sqrt{2}}{160} = \frac{\sqrt{2}}{20} \approx 0.1 ]

This approximates to $2^m \times 5^n$ where $m=1$ and $n=1$ .

Step 3

Explain why \int_{6.3}^{6.4} 2x^3 \, dx is not a suitable approximation for \int_{6.3}^{6.4} (sin x - sin x \cos 2x) \, dx

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Answer

The limits for the integral of $sin x - sin x \cos 2x$ from $6.3$ to $6.4$ are too large for using the small angle approximation. The approximation fails because the values of $x$ in this range are not small, hence the behavior of the original function is significantly different from the polynomial approximation given by $2x^3$ .

Step 4

Explain how \int_{6.3}^{6.4} (sin x - sin x \cos 2x) \, dx may be approximated by \int_{a}^{b} 2x^3 \, dx

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Answer

The limits can be changed to either evaluate over intervals where $sin x$ is dominant or utilize periodicity. For instance, the substitution of the limits could yield:

$\int_{6.3 - 2\pi}^{6.4 - 2\pi}$ or any equivalent form that repeats for sinusoidal functions. This allows us to find suitable values $a$ and $b$ such that both integrals approximate the same area, exploiting the periodic nature of the sine function.

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