FP2 June 2006 Q5
5.
(a) Find the Taylor expansion of \(\cos 2x\) in ascending powers of \(\left(x - \dfrac{\pi}{4}\right)\) up to and including the term in \(\left(x - \dfrac{\pi}{4}\right)^5\). (5)
(b) Use your answer to (a) to obtain an estimate of \(\cos 2\), giving your answer to 6 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \cos 2x, \qquad \mathrm{f}\left(\tfrac{\pi}{4}\right) = 0\) | |
| \(\mathrm{f}'(x) = -2\sin 2x, \qquad \mathrm{f}'\left(\tfrac{\pi}{4}\right) = -2\) | M1 |
| \(\mathrm{f}''(x) = -4\cos 2x, \qquad \mathrm{f}''\left(\tfrac{\pi}{4}\right) = 0\) | |
| \(\mathrm{f}'''(x) = 8\sin 2x, \qquad \mathrm{f}'''\left(\tfrac{\pi}{4}\right) = 8\) | A1 |
| \(\mathrm{f}^{(\mathrm{iv})}(x) = 16\cos 2x, \qquad \mathrm{f}^{(\mathrm{iv})}\left(\tfrac{\pi}{4}\right) = 0\) | |
| \(\mathrm{f}^{(\mathrm{v})}(x) = 32\sin 2x, \qquad \mathrm{f}^{(\mathrm{v})}\left(\tfrac{\pi}{4}\right) = -32\) | A1 |
| \(\cos 2x = \mathrm{f}\left(\tfrac{\pi}{4}\right) + \mathrm{f}'\left(\tfrac{\pi}{4}\right)\left(x - \tfrac{\pi}{4}\right) + \dfrac{\mathrm{f}''\left(\frac{\pi}{4}\right)}{2}\left(x - \tfrac{\pi}{4}\right)^2 + \dfrac{\mathrm{f}'''\left(\frac{\pi}{4}\right)}{3!}\left(x - \tfrac{\pi}{4}\right)^3 + \ldots\) Three terms are sufficient to establish method | M1 |
| \(\cos 2x = -2\left(x - \tfrac{\pi}{4}\right) + \tfrac{4}{3}\left(x - \tfrac{\pi}{4}\right)^3 - \tfrac{4}{15}\left(x - \tfrac{\pi}{4}\right)^5 + \ldots\) | A1 |
| (5) |
Notes
(corrected from the printed mark scheme: the \(\left(x - \tfrac{\pi}{4}\right)^3\) term of the general expansion is printed with \(\mathrm{f}'\left(\tfrac{\pi}{4}\right)\) instead of \(\mathrm{f}'''\left(\tfrac{\pi}{4}\right)\))
| Scheme | Marks |
|---|---|
| Substitute \(x = 1\) \(\left(1 - \tfrac{\pi}{4} \approx 0.21460\right)\) | B1 |
| \(\cos 2 = -2\left(x - \tfrac{\pi}{4}\right) + \tfrac{4}{3}\left(x - \tfrac{\pi}{4}\right)^3 - \tfrac{4}{15}\left(x - \tfrac{\pi}{4}\right)^5 + \ldots\) \(\approx -0.416147\) cao | M1 A1 |
| (3) | |
| (8 marks) |