A2 June 2024 Paper 1 Q4
4. The complex number \(z = \mathrm{e}^{\mathrm{i}\theta}\), where \(\theta\) is real.
| Scheme | Marks | AO |
|---|---|---|
| \(z^n + \dfrac{1}{z^n} \equiv \mathrm{e}^{\mathrm{i}n\theta} + \dfrac{1}{\mathrm{e}^{\mathrm{i}n\theta}} \equiv \mathrm{e}^{\mathrm{i}n\theta} + \mathrm{e}^{-\mathrm{i}n\theta}\) | M1 | 1.1b |
| \(\equiv \cos n\theta + \mathrm{i}\sin n\theta + \cos n\theta - \mathrm{i}\sin n\theta \equiv 2\cos n\theta\) * | A1* | 2.1 |
| (2) |
Notes
M1: Substitutes \(z\) into the LHS and simplifies the powers as shown. Allow if they go direct to trigonometric expressions without exponentials. The mark is for sorting out the negative index.
A1*: Converts the exponential form to trigonometric form correctly and correctly completes the proof with no errors seen. The trigonometric expansion must be clearly seen. Condone missing brackets in e.g \(\cos -n\theta\) terms if intent is clear. Note the LHS of the identity may be implied.
Alt (a)
| Scheme | Marks | AO |
|---|---|---|
| \(z^n + \dfrac{1}{z^n} = \cos n\theta + \mathrm{i}\sin n\theta + \dfrac{1}{\cos n\theta + \mathrm{i}\sin n\theta} = \dfrac{\cos^2 n\theta + 2\mathrm{i}\cos n\theta\sin n\theta - \sin^2 n\theta + 1}{\cos n\theta + \mathrm{i}\sin n\theta}\) | M1 | 1.1b |
| \(= \dfrac{2\cos^2 n\theta + 2\mathrm{i}\cos n\theta\sin n\theta}{\cos n\theta + \mathrm{i}\sin n\theta} = \dfrac{2\cos n\theta(\cos n\theta + \mathrm{i}\sin n\theta)}{\cos n\theta + \mathrm{i}\sin n\theta} = 2\cos n\theta\) * | A1* | 2.1 |
| (2) |
(corrected from the printed mark scheme: the denominator in the first line is printed as \(\cos n\theta + \mathrm{i}\sin\theta\))
Alt: by De Moivre
M1: Applies De Moivre on both terms and puts over a common denominator.
A1*: Complete correctly, using \(1 - \sin^2 n\theta = \cos^2 n\theta\) and cancelling \(\cos n\theta + \mathrm{i}\sin n\theta\). No errors seen.
| Scheme | Marks | AO |
|---|---|---|
| \(\left(z + z^{-1}\right)^5 = 32\cos^5\theta\) | B1 | 2.2a |
| \(\left(z + z^{-1}\right)^5 = z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}\) | M1 A1 | 1.1b 1.1b |
| \(32\cos^5\theta = \left(z^5 + z^{-5}\right) + 5\left(z^3 + z^{-3}\right) + 10\left(z + z^{-1}\right)\) \(= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta\) | M1 | 2.1 |
| \(\cos^5\theta = \dfrac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)\) * | A1* | 1.1b |
| (5) |
Notes
B1: Deduces that \(\left(z + z^{-1}\right)^5 = 32\cos^5\theta\) Do not accept \(2^5\) for 32. May be implied.
M1: Attempts to expand \(\left(z + z^{-1}\right)^5\). Correct binomial coefficients must be used, terms need not be simplified. Condone at most one slip in powers.
A1: Correct expansion, terms need not be gathered but powers must have been simplified.
M1: Sets their expressions equal and applies the result from (a) – grouping must be shown.
A1*: Reaches the printed answer with no errors and relevant steps all shown.
Alt (b)
| Scheme | Marks | AO |
|---|---|---|
| \(\cos 5\theta = \mathrm{Re}(\cos 5\theta + \mathrm{i}\sin 5\theta) = \mathrm{Re}(\cos\theta + \mathrm{i}\sin\theta)^5\) | B1 | 2.2a |
| \((\cos\theta + \mathrm{i}\sin\theta)^5 = c^5 + 5\mathrm{i}c^4s + 10\mathrm{i}^2c^3s^2 + 10\mathrm{i}^3c^2s^3 + 5\mathrm{i}^4cs^4 + \mathrm{i}^5s^5\) | M1 | 1.1b |
| \(\mathrm{Re}(\cos\theta + \mathrm{i}\sin\theta)^5 = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\) | A1 | 1.1b |
| \(\cos 5\theta = \cos^5\theta - 10\cos^3\theta\left(1 - \cos^2\theta\right) + 5\cos\theta\left(1 - \cos^2\theta\right)^2\) \(= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta\) \(= 16\cos^5\theta - \dfrac{20}{4}(\cos 3\theta + 3\cos\theta) + 5\cos\theta\) | M1 | 2.1 |
| \(\Rightarrow \cos^5\theta = \dfrac{1}{16}(\cos 5\theta + 5\cos 3\theta + 10\cos\theta)\) * | A1* | 1.1b |
| (5) |
Alt: by De Moivre
B1: Correctly stated or clearly implied De Moivre statement for \(\cos 5\theta\)
M1: Attempts to expand \((\cos\theta + \mathrm{i}\sin\theta)^5\) Correct coefficients but allow one slip per main scheme. The powers of i need not be simplified for the attempt at expansion, accept if only the real terms are shown. Allow \(c\) and \(s\) notation.
A1: Correct real terms extracted with the i's removed.
M1: Applies \(\sin^2\theta = 1 - \cos^2\theta\) to reduce to an equation in \(\cos\theta\) and applies \(\cos^3\theta = \dfrac{1}{4}(\cos 3\theta + 3\cos\theta)\) (quoted or derived – allow a slip if derived) to get to an equation without powers of cos terms.
A1*: Reaches the printed answer with no errors and relevant steps all shown.
| Scheme | Marks | AO |
|---|---|---|
| \(\cos 5\theta + 5\cos 3\theta + 10\cos\theta = -2\cos\theta \Rightarrow 16\cos^5\theta = -2\cos\theta\) | B1 | 3.1a |
| \(2\cos\theta\left(8\cos^4\theta + 1\right) = 0 \Rightarrow \theta = \ldots\) | M1 | 1.1b |
| \(8\cos^4\theta + 1 = 0\) has no solution so \(\cos\theta = 0\) \(\theta = \dfrac{\pi}{2},\ \dfrac{3\pi}{2}\) | A1 | 2.2a |
| (3) | ||
| (10 marks) |
Notes
B1: Uses the result from (b) to deduce the correct equation.
M1: Must have attempted to use part (b) to obtain \(\alpha\cos^5\theta = \beta\cos\theta\) or equivalent. Collects to one side and attempts to factorise and solve. Note dividing through by \(\cos\theta\) is M0.
A1: Rejects the inappropriate solution and selects \(\cos\theta = 0\) and obtains the correct values only. The equation must have been correct. There must have been some consideration of the \(8\cos^4\theta + 1\) e.g. stating \(8\cos^4\theta \gt 0\) so no solutions, or attempting to find complex roots and deducing no answers. May be minimal, but some consideration that no roots arise from this part must have been given.
Note: The correct answer will appear from incorrect attempts – the M must be gained in order to award the A. E.g. assuming the equation reduces to \(\cos^5\theta = 0\) will score B0M0A0.
Likewise, answers only scores B0M0A0 (questions says hence so use of (b) must be seen).