A2 October 2021 Paper 1 Q5
5 Use de Moivre’s theorem to find the constants \(A\), \(B\) and \(C\) in the identity
\(\sin^5\theta \equiv A\sin\theta + B\sin 3\theta + C\sin 5\theta\). [4]
| Scheme | Marks | AO |
|---|---|---|
| Take \(z = \cos\theta + \mathrm{i}\sin\theta \Rightarrow z^{-1} = \cos\theta - \mathrm{i}\sin\theta\) \(\Rightarrow z - \dfrac{1}{z} = 2\mathrm{i}\sin\theta\) | M1 | 2.1 |
| \(\left(z - \dfrac{1}{z}\right)^5 = 32\mathrm{i}\sin^5\theta = z^5 - 5z^3 + 10z - \dfrac{10}{z} + \dfrac{5}{z^3} - \dfrac{1}{z^5}\) | A1 | 1.1 |
| \(\Rightarrow 32\mathrm{i}\sin^5\theta = \left(z^5 - \dfrac{1}{z^5}\right) - 5\left(z^3 - \dfrac{1}{z^3}\right) + 10\left(z - \dfrac{1}{z}\right)\) \(= 2\mathrm{i}\sin 5\theta - 10\mathrm{i}\sin 3\theta + 20\mathrm{i}\sin\theta\) \(\Rightarrow \sin^5\theta = \dfrac{5}{8}\sin\theta - \dfrac{5}{16}\sin 3\theta + \dfrac{1}{16}\sin 5\theta\). | M1 | 1.1 |
| i.e. \(A = \dfrac{5}{8},\ B = -\dfrac{5}{16},\ C = \dfrac{1}{16}\) | A1 | 2.2a |
| [4] |
Notes
M1: Use of \(z\) and de Moivre Sight of i is necessary
A1: Both sides
M1: Attempt conversion into sin soi
A1: All three stated
Alternative method 1
| Scheme | Marks |
|---|---|
| \((\sin\theta)^5 = \left(\dfrac{\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}}{2\mathrm{i}}\right)^5\) | M1 |
| \(= \dfrac{1}{(2\mathrm{i})^5}\left(\mathrm{e}^{5\mathrm{i}\theta} - 5\mathrm{e}^{3\mathrm{i}\theta} + 10\mathrm{e}^{\mathrm{i}\theta} - 10\mathrm{e}^{-\mathrm{i}\theta} + 5\mathrm{e}^{-3\mathrm{i}\theta} - \mathrm{e}^{-5\mathrm{i}\theta}\right)\) | A1 |
| \(= \dfrac{1}{32\mathrm{i}}\left(\left(\mathrm{e}^{5\mathrm{i}\theta} - \mathrm{e}^{-5\mathrm{i}\theta}\right) - 5\left(\mathrm{e}^{3\mathrm{i}\theta} - \mathrm{e}^{-3\mathrm{i}\theta}\right) + 10\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)\right)\) \(= \dfrac{1}{16}(\sin 5\theta - 5\sin 3\theta + 10\sin\theta)\) \(\Rightarrow \sin^5\theta = \dfrac{5}{8}\sin\theta - \dfrac{5}{16}\sin 3\theta + \dfrac{1}{16}\sin 5\theta\) | M1 |
| \(\Rightarrow A = \dfrac{5}{8},\ B = -\dfrac{5}{16},\ C = \dfrac{1}{16}\) | A1 |
M1: Sight of i in the denominator is necessary
M1: Collection to convert back
A1: All three stated
Alternative method 2
| Scheme | Marks |
|---|---|
| \(z = \cos\theta + i\sin\theta\) \(\Rightarrow z^5 = \cos 5\theta + i\sin 5\theta\) \(= (\cos\theta + i\sin\theta)^5\) \(= \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta\) \(\Rightarrow \sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\) \(= 5\left(1 - \sin^2\theta\right)^2\sin\theta - 10\left(1 - \sin^2\theta\right)\sin^3\theta + \sin^5\theta\) \(= 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta\) \(= 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta\) | M1 |
| \(z^3 = \cos 3\theta + i\sin 3\theta\) \(= (\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta\) \(\Rightarrow \sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta\) \(= 3\sin\theta - 4\sin^3\theta\) | A1 |
| \(\Rightarrow \sin 5\theta - 5\sin 3\theta = -10\sin\theta + 16\sin^5\theta\) \(\Rightarrow 16\sin^5\theta = 10\sin\theta - 5\sin 3\theta + \sin 5\theta\) | M1 |
| \(\Rightarrow A = \dfrac{10}{16} = \dfrac{5}{8},\ B = -\dfrac{5}{16},\ C = \dfrac{1}{16}\) | A1 |
| [4] |
M1: De Moivre
A1: for both
M1: Eliminate \(\sin^3\theta\)
A1: All three stated
(Corrected from the printed mark scheme: Alternative method 2 is printed with \(B = -\dfrac{5}{8}\) and with \(z^3 = (\cos\theta + i\sin\theta)^5\); these are \(B = -\dfrac{5}{16}\) and \((\cos\theta + i\sin\theta)^3\), as typed above.)