A2 June 2021 Paper 2 Q13
13
Find the other solutions to the equation \(\cos 6\theta = 0\) for \(0 \leqslant \theta \leqslant \pi\) [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains at least two correct solutions | M1 | 1.1a |
| Obtains all correct solutions | A1 | 1.1b |
| (2) |
Typical solution
\[6\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}\]\[\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}\](in addition to given solutions)
| Scheme | Marks | AO |
|---|---|---|
| Expands \((\cos\theta + \mathrm{i}\sin\theta)^6\) | M1 | 1.1a |
| Equates real parts | M1 | 3.1a |
| Obtains correct expression for real part in terms of powers of \(\cos\theta\) and \(\sin\theta\) | A1 | 1.1b |
| Uses trig identity to express real part in terms of \(\cos\theta\) | M1 | 3.1a |
| Completes a rigorous argument to obtain the required result | R1 | 2.1 |
| (5) |
Typical solution
By de Moivre’s theorem
\[\cos 6\theta + \mathrm{i}\sin 6\theta = (\cos\theta + \mathrm{i}\sin\theta)^6\]\[\begin{aligned} = \cos^6\theta &+ 6\cos^5\theta\,(\mathrm{i}\sin\theta) + 15\cos^4\theta\,(\mathrm{i}\sin\theta)^2 \\ &+ 20\cos^3\theta\,(\mathrm{i}\sin\theta)^3 + 15\cos^2\theta\,(\mathrm{i}\sin\theta)^4 \\ &+ 6\cos\theta\,(\mathrm{i}\sin\theta)^5 + (\mathrm{i}\sin\theta)^6 \end{aligned}\]Equating real parts
\[\cos 6\theta = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta\]\[\begin{aligned} &= \cos^6\theta - 15\cos^4\theta\,(1 - \cos^2\theta) + 15\cos^2\theta\,(1 - \cos^2\theta)^2 - (1 - \cos^2\theta)^3 \\ &= \cos^6\theta - 15\cos^4\theta + 15\cos^6\theta + 15\cos^2\theta - 30\cos^4\theta \\ &\qquad + 15\cos^6\theta - 1 + 3\cos^2\theta - 3\cos^4\theta + \cos^6\theta \end{aligned}\]\[\cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1\]| Scheme | Marks | AO |
|---|---|---|
| Uses the fact that either \(\theta = \frac{\pi}{4}\) or \(\theta = \frac{3\pi}{4}\) is a solution to the first equation to deduce that it is also a solution to the second equation | M1 | 2.2a |
| Uses the factor theorem | M1 | 3.1a |
| Multiplies their linear factors together | M1 | 1.1a |
| Obtains the correct result (oe) | R1 | 2.1 |
| (4) |
Typical solution
\[\theta = \frac{\pi}{4} \Rightarrow \cos 6\theta = 0\]\[\Rightarrow 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1 = 0\]from part (b)
\(\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\) \(\quad\therefore \left(\cos\theta - \frac{1}{\sqrt{2}}\right)\) is a factor of \(32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1\)
Similarly \(\cos\frac{3\pi}{4} = -\frac{1}{\sqrt{2}}\) and \(\left(\cos\theta + \frac{1}{\sqrt{2}}\right)\) is also a factor of the expression.
So \(\left(\cos\theta - \frac{1}{\sqrt{2}}\right)\left(\cos\theta + \frac{1}{\sqrt{2}}\right) = \left(\cos^2\theta - \frac{1}{2}\right)\) is a factor and \((2\cos^2\theta - 1)\) is a factor
| Scheme | Marks | AO |
|---|---|---|
| Divides the polynomial by their quadratic factor | M1 | 3.1a |
| Solves their quartic equation as a quadratic in \(c^2\) | M1 | 1.1a |
| Explains that the roots of the quartic correspond to the cosines of the angles found in part (a) | E1 | 2.4 |
| Obtains all correct roots of the quartic | A1 | 1.1b |
| Uses a rigorous argument to obtain the required result, including a reason why that particular root corresponds to \(\cos\left(\dfrac{11\pi}{12}\right)\) | R1 | 2.1 |
| (5) | ||
| (16 marks) |
Typical solution
Let \(c = \cos\theta\)
\[32c^6 - 48c^4 + 18c^2 - 1 = (2c^2 - 1)(16c^4 - 16c^2 + 1)\]For \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\), \(2c^2 - 1 = 0\)
So for the other four roots, which are the cosines of the other four solutions to \(\cos 6\theta = 0\),
\[16c^4 - 16c^2 + 1 = 0\]Solving this as a quadratic in \(c^2\) gives
\[c^2 = \frac{2 \pm \sqrt{3}}{4}\]So
\[c = \pm\sqrt{\frac{2 \pm \sqrt{3}}{4}}\]Of the angles \(\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}\), and \(\frac{11\pi}{12}\),
\(\frac{11\pi}{12}\) has the negative cosine of the greatest magnitude
Therefore \(\cos\left(\frac{11\pi}{12}\right) = -\sqrt{\frac{2 + \sqrt{3}}{4}}\)