A2 June 2024 Paper 1 Q10
10 The complex numbers \(z\) and \(w\) are defined by
\[z = \cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}\]and
\[w = \cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6}\]By evaluating the product \(zw\), show that
\[\tan\frac{5\pi}{12} = 2 + \sqrt{3}\][6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms the product \(zw\) | M1 | 1.1a |
| Obtains \(\dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} + \mathrm{i}\left(\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}\right)\) | A1 | 1.1b |
| States or uses \(\arg(zw) = \arg(z) + \arg(w)\) | M1 | 3.1a |
| States \(\dfrac{\pi}{4} + \dfrac{\pi}{6} = \dfrac{5\pi}{12}\) | B1 | 1.1b |
| Uses their \(zw = \dfrac{\sqrt{6}}{4} - \dfrac{\sqrt{2}}{4} + \mathrm{i}\left(\dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}\right)\) to deduce an expression for \(\tan\dfrac{5\pi}{12}\) | M1 | 2.2a |
| Completes a reasoned argument to obtain \(\tan\dfrac{5\pi}{12} = 2 + \sqrt{3}\) AG | R1 | 2.1 |
| (6 marks) |
Typical solution
\[\begin{aligned}zw &= \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\mathrm{i}\right)\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i}\right) \\ &= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} + \mathrm{i}\left(\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}\right)\end{aligned}\]Also
\[\begin{aligned}zw &= \cos\left(\frac{\pi}{4} + \frac{\pi}{6}\right) + \mathrm{i}\sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) \\ &= \cos\frac{5\pi}{12} + \mathrm{i}\sin\frac{5\pi}{12}\end{aligned}\]\[\tan\frac{5\pi}{12} = \frac{\sin\frac{5\pi}{12}}{\cos\frac{5\pi}{12}} = \frac{\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}}{\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}} = 2 + \sqrt{3}\]