FP2 June 2011 Q7
7.
(a) Use de Moivre’s theorem to show that \[\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\] (5)
Hence, given also that \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\),
(b) find all the solutions of \[\sin 5\theta = 5\sin 3\theta,\] in the interval \(0 \leqslant \theta < 2\pi\). Give your answers to 3 decimal places. (6)
| Scheme | Marks |
|---|---|
| \(\sin 5\theta = \mathrm{Im}(\cos\theta + \mathrm{i}\sin\theta)^5\) | B1 |
| \(5\cos^4\theta(\mathrm{i}\sin\theta) + 10\cos^2\theta(\mathrm{i}^3\sin^3\theta) + \mathrm{i}^5\sin^5\theta\) | M1 |
| \(= \mathrm{i}(5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta)\) | A1 |
| \(\left(\mathrm{Im}(\cos\theta + \mathrm{i}\sin\theta)^5\right) = 5\sin\theta(1 - \sin^2\theta)^2 - 10\sin^3\theta(1 - \sin^2\theta) + \sin^5\theta\) | M1 |
| \(\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\) (*) | A1cso |
| (5) |
Notes
Award B if solution considers Imaginary parts and equates to \(\sin 5\theta\)
1st M1 for correct attempt at expansion and collection of imaginary parts
2nd M1 for substitution powers of \(\cos\theta\)
| Scheme | Marks |
|---|---|
| \(16\sin^5\theta - 20\sin^3\theta + 5\sin\theta = 5(3\sin\theta - 4\sin^3\theta)\) | M1 |
| \(16\sin^5\theta - 10\sin\theta = 0\) | M1 |
| \(\sin^4\theta = \dfrac{5}{8} \qquad \theta = 1.095\) | A1 |
| Inclusion of solutions from \(\sin\theta = -\sqrt[4]{\dfrac{5}{8}}\) | M1 |
| Other solutions: \(\theta = 2.046,\ 4.237,\ 5.188\) | A1 |
| \(\sin\theta = 0 \Rightarrow \theta = 0,\ \theta = \pi\ (3.142)\) | B1 |
| (6) | |
| (11 marks) |
Notes
1st M for substituting correct expressions
2nd M for attempting to form equation
Imply 3rd M if 4.237 or 5.188 seen. Award for their negative root.
Ignore \(2\pi\) but 2nd A0 if other extra solutions given.