June 2022 Paper 3 Q12
12
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
The relevant parts of the article “Approximating the sine function” are reproduced below; the line numbers are those printed on the Insert.
Lines 23–24
The approximation \(\sin x \approx \dfrac{16x(\pi - x)}{5\pi^2 - 4x(\pi - x)}\) was discovered by an Indian mathematician named Bhaskara in the 7th century.Lines 29–31
The percentage error in approximating \(\sin x\) by \(\dfrac{16x(\pi - x)}{5\pi^2 - 4x(\pi - x)}\) is less than 2% throughout the interval \(0 \leqslant x \leqslant \pi\). The Bhaskara approximation for \(\sin x\) can be used to derive the following approximation for \(\cos x\); \(\cos x \approx \dfrac{\pi^2 - 4x^2}{\pi^2 + x^2}\).
| Scheme | Marks | AO |
|---|---|---|
| \(y = \sin\left(x + \frac{\pi}{2}\right)\) is a translation of \(y = \sin x\) | M1 | 3.1a |
| \(\dfrac{\pi}{2}\) to left and so is the same as \(y = \cos x\) | E1 | 2.2a |
| [2] |
Notes
M1: OR \(\sin\left(x + \frac{\pi}{2}\right) = \sin x\cos\frac{\pi}{2} + \sin\frac{\pi}{2}\cos x\)
E1: OR \(\sin\left(x + \frac{\pi}{2}\right) = \sin x \times 0 + 1 \times \cos x = \cos x\)
Convincing completion (AG)
| Scheme | Marks | AO |
|---|---|---|
| \([\cos x \approx]\ \dfrac{16\left(x + \frac{\pi}{2}\right)\left(\frac{\pi}{2} - x\right)}{5\pi^2 - 4\left(x + \frac{\pi}{2}\right)\left(\frac{\pi}{2} - x\right)}\) | M1* | 2.1 |
| \([\cos x \approx]\ \dfrac{16\left(\frac{\pi^2}{4} - x^2\right)}{5\pi^2 - 4\left(\frac{\pi^2}{4} - x^2\right)}\) | DM1 | 1.1 |
| \([\cos x \approx]\ \dfrac{4\pi^2 - 16x^2}{4\pi^2 + 4x^2} = \dfrac{\pi^2 - 4x^2}{\pi^2 + x^2}\) | A1 | 1.1 |
| [3] |
Notes
M1*: Substitute \(x + \dfrac{\pi}{2}\)
DM1: Multiplying out brackets (dep on first M1)
A1: Convincing completion (AG)
Check continuation page