C3 June 2009 Q6
6.
The curves \(C_1\) and \(C_2\) have equations
\[\begin{aligned} C_1&:\ y = 3\sin 2x \\ C_2&:\ y = 4\sin^2 x - 2\cos 2x \end{aligned}\]| Scheme | Marks |
|---|---|
| \(A = B \Rightarrow \cos(A + A) = \cos 2A = \underline{\cos A\cos A - \sin A\sin A}\) | M1 |
| \(\cos 2A = \cos^2 A - \sin^2 A\) and \(\cos^2 A + \sin^2 A = 1\) gives | |
| \(\underline{\cos 2A = 1 - \sin^2 A - \sin^2 A = 1 - 2\sin^2 A}\) (as required) | A1 AG |
| (2) |
Notes
M1: Applies \(A = B\) to \(\cos(A + B)\) to give the underlined equation or \(\cos 2A = \underline{\cos^2 A - \sin^2 A}\)
A1 AG: Complete proof, with a link between LHS and RHS. No errors seen.
| Scheme | Marks |
|---|---|
| \(C_1 = C_2 \Rightarrow 3\sin 2x = 4\sin^2 x - 2\cos 2x\) | M1 |
| \(3\sin 2x = 4\left(\dfrac{1 - \cos 2x}{2}\right) - 2\cos 2x\) | M1 |
| \(3\sin 2x = 2(1 - \cos 2x) - 2\cos 2x\) \(3\sin 2x = 2 - 2\cos 2x - 2\cos 2x\) | |
| \(3\sin 2x + 4\cos 2x = 2\) | A1 AG |
| (3) |
Notes
M1: Eliminating \(y\) correctly.
M1: Using result in part (a) to substitute for \(\sin^2 x\) as \(\dfrac{\pm 1 \pm \cos 2x}{2}\) or \(k\sin^2 x\) as \(k\left(\dfrac{\pm 1 \pm \cos 2x}{2}\right)\) to produce an equation in only double angles.
A1 AG: Rearranges to give correct result
| Scheme | Marks |
|---|---|
| \(3\sin 2x + 4\cos 2x = R\cos(2x - \alpha)\) \(3\sin 2x + 4\cos 2x = R\cos 2x\cos\alpha + R\sin 2x\sin\alpha\) | |
| Equate \(\sin 2x\): \(3 = R\sin\alpha\) Equate \(\cos 2x\): \(4 = R\cos\alpha\) | |
| \(R = \sqrt{3^2 + 4^2};= \sqrt{25} = 5\) | B1 |
| \(\tan\alpha = \tfrac{3}{4} \Rightarrow \alpha = 36.86989765\ldots^\circ\) | M1 A1 |
| Hence, \(3\sin 2x + 4\cos 2x = 5\cos(2x - 36.87)\) | |
| (3) |
Notes
B1: \(R = 5\)
M1: \(\tan\alpha = \pm\tfrac{3}{4}\) or \(\tan\alpha = \pm\tfrac{4}{3}\) or \(\sin\alpha = \pm\frac{3}{\text{their } R}\) or \(\cos\alpha = \pm\frac{4}{\text{their } R}\)
A1: awrt 36.87
| Scheme | Marks |
|---|---|
| \(3\sin 2x + 4\cos 2x = 2\) \(5\cos(2x - 36.87) = 2\) | |
| \(\cos(2x - 36.87) = \dfrac{2}{5}\) | M1 |
| \((2x - 36.87) = 66.42182\ldots^\circ\) | A1 |
| \((2x - 36.87) = 360 - 66.42182\ldots^\circ\) | |
| Hence, \(x = 51.64591\ldots^\circ,\ 165.22409\ldots^\circ\) | A1 A1 |
| (4) | |
| (12 marks) |
Notes
M1: \(\cos(2x \pm \text{their } \alpha) = \dfrac{2}{\text{their } R}\)
A1: awrt 66
A1: One of either awrt 51.6 or awrt 51.7 or awrt 165.2 or awrt 165.3
A1: Both awrt 51.6 AND awrt 165.2
If there are any EXTRA solutions inside the range \(0 \leqslant x \lt 180^\circ\) then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range \(0 \leqslant x \lt 180^\circ\).