A2 June 2020 Paper 2 Q9
9 The matrix \(\mathbf{C} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}\), where \(a\) and \(b\) are positive real numbers,
and \(\mathbf{C}^2 = \begin{bmatrix} \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \\[1ex] \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{bmatrix}\)
Use \(\mathbf{C}\) to show that \(\cos\dfrac{\pi}{12}\) can be written in the form \(\dfrac{\sqrt{\sqrt{m} + n}}{2}\), where \(m\) and \(n\) are integers. [7 marks]
| Scheme | Marks | AO |
|---|---|---|
| Explains that \(\mathbf{C}^2\) represents a rotation of \(\dfrac{\pi}{6}\) or that \(\mathbf{C}\) represents a rotation of \(\dfrac{\pi}{12}\) | E1 | 2.4 |
| Squares matrix \(\mathbf{C}\) with at least two correct elements. | M1 | 3.1a |
| Forms two simultaneous equations in \(a\) and \(b\) using their squared \(\mathbf{C}\) and the given \(\mathbf{C}^2\) | M1 | 1.1a |
| Forms two correct simultaneous equations. | A1 | 1.1b |
| Eliminates \(a\) or \(b\) and forms a quadratic equation in \(b^2\) or \(a^2\) | M1 | 1.1a |
| Finds the correct value of \(a^2 = \dfrac{\sqrt{3} + 2}{4}\) | A1 | 1.1b |
| Completes a rigorous argument to show that \(\cos\dfrac{\pi}{12} = \dfrac{\sqrt{\sqrt{3} + 2}}{2}\) by explaining that \(\mathbf{C}\) represents a rotation of \(\dfrac{\pi}{12}\) | R1 | 2.1 |
| (7 marks) |
Typical solution
\[\begin{bmatrix} a & -b \\ b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2 - b^2 & -2ab \\ 2ab & a^2 - b^2 \end{bmatrix}\]\[2ab = \frac{1}{2}\]\[a^2 - b^2 = \frac{\sqrt{3}}{2}\]\[b = \frac{1}{4a}\]\[a^2 - \frac{1}{16a^2} = \frac{\sqrt{3}}{2}\]\[16a^4 - 8\sqrt{3}a^2 - 1 = 0\]\[a^2 = \frac{\sqrt{3} + 2}{4}\]\[a = \frac{\sqrt{\sqrt{3} + 2}}{2} \quad \text{since } a \gt 0\]\(\mathbf{C}^2\) represents a rotation of \(\dfrac{\pi}{6}\), therefore \(\mathbf{C}\) represents a rotation of \(\dfrac{1}{2}\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{12}\)
So \(a = \cos\dfrac{\pi}{12}\) and \(\cos\dfrac{\pi}{12} = \dfrac{\sqrt{\sqrt{3} + 2}}{2}\)