A2 June 2019 Paper 1 Q8
8
(a) If \(z = \cos\theta + \mathrm{i}\sin\theta\), use de Moivre’s theorem to prove that\[z^n - \frac{1}{z^n} = 2\mathrm{i}\sin n\theta\] [3 marks]
(b) Express \(\sin^5\theta\) in terms of \(\sin 5\theta\), \(\sin 3\theta\) and \(\sin\theta\) [4 marks]
(c) Hence show that\[\int_0^{\frac{\pi}{3}} \sin^5\theta \,\mathrm{d}\theta = \frac{53}{480}\] [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains correct expression for \(z^n\) in terms of \(\cos n\theta\) and \(\sin n\theta\) | B1 | 1.1b |
| Obtains correct expression for \(\dfrac{1}{z^n}\) in terms of \(\cos n\theta\) and \(\sin n\theta\) Or expresses whole LHS as \(\dfrac{-2\sin^2 n\theta + 2\mathrm{i}\cos n\theta\sin n\theta}{\cos n\theta + \mathrm{i}\sin n\theta}\) | B1 | 1.1b |
| Completes a rigorous argument (with all intermediate steps) to show the required result, using properties of sine and cosine functions to obtain results in terms of \(\cos n\theta\) and \(\sin n\theta\) | R1 | 2.1 |
Typical solution
\[\begin{aligned} z^n &= (\cos\theta + \mathrm{i}\sin\theta)^n \\ &= \cos n\theta + \mathrm{i}\sin n\theta \end{aligned}\]\[\begin{aligned} \frac{1}{z^n} &= (\cos\theta + \mathrm{i}\sin\theta)^{-n} \\ &= \cos(-n\theta) + \mathrm{i}\sin(-n\theta) \\ &= \cos n\theta - \mathrm{i}\sin n\theta \end{aligned}\]\[\begin{aligned} z^n - \frac{1}{z^n} &= \cos n\theta + \mathrm{i}\sin n\theta - (\cos n\theta - \mathrm{i}\sin n\theta) \\ &= 2\mathrm{i}\sin n\theta \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Selects the correct process by expanding \(\left(z - \dfrac{1}{z}\right)^5\) | M1 | 3.1a |
| Obtains three pairs of terms in the form \(z^n - \dfrac{1}{z^n}\) (ignore LHS) | M1 | 1.1a |
| Replaces each pair with the correct \(\sin n\theta\) (ignore coefficients and LHS) | M1 | 1.1a |
| Obtains correct result | A1 | 1.1b |
Typical solution
\[\left(z - \frac{1}{z}\right)^5 = z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5}\]\[(2\mathrm{i}\sin\theta)^5 = z^5 - z^{-5} - 5\left(z^3 - z^{-3}\right) + 10\left(z - z^{-1}\right)\]\[32\mathrm{i}\sin^5\theta = 2\mathrm{i}\sin 5\theta - 5(2\mathrm{i}\sin 3\theta) + 10(2\mathrm{i}\sin\theta)\]\[\sin^5\theta = \frac{1}{16}\sin 5\theta - \frac{5}{16}\sin 3\theta + \frac{5}{8}\sin\theta\]| Scheme | Marks | AO |
|---|---|---|
| Integrates their answer to part (b) correctly, provided all terms in integrand are of the form \(k\sin n\theta\) | M1 | 1.1a |
| Shows substitution clearly | M1 | 2.4 |
| Completes a rigorous argument to show the required result. NMS = 0/3 | R1 | 2.1 |
| (10 marks) |