| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve double/multiple angle equation |
| Difficulty | Moderate -0.8 This is a straightforward C2 trigonometry question requiring basic graph reading, knowledge of transformations, and solving a standard trig equation using a calculator. Part (a) is trivial coordinate reading, part (b) tests understanding of horizontal stretch (standard bookwork), and part (c) is a routine inverse cosine calculation with multiple solutions in the given range. All parts are below average difficulty for A-level, requiring only recall and standard procedures with no problem-solving insight. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(A(0,1)\); \(B(45°,0)\); \(C(270°,-1)\) | B1, B1, B1, B1 | Condone radians; Condone (0.785,0) or better.; B1 for 270; B1 for −1 |
| Total for 7(a): 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Stretch (I) in x-direction (II) with a scale factor \(\frac{1}{2}\) (III) | M1A1 | More than one transformation is M0; M1 for (I) and either (II) or (III) |
| Total for 7(b): 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos^{-1}0.37 = "68.284..."\) (\(=\alpha\)); \(x = \frac{\alpha}{2} = 34.1(42.)\); \(x = 180 - \frac{\alpha}{2}\); \(x = 180 + \frac{\alpha}{2}\) and \(x = 180 + 180 - \frac{\alpha}{2}\); \(2x = 68.284...;291.715...;\); \(428.284...;651.715...\); \(x = (34.1;)\); \(145.9;214.1;325.9\) | M1, A1, m1, m1, A1 | Cos\(^{-1}\)(0.37) (PI eg 68.3 or 1.19); Condone 34.2°, 34° or 0.596 rads; OE eg \(2x = 360 - \alpha\); OE Need both (OE for \(2x =\) ) with no extras (quadrants) within the given interval; Dep. on all three method marks. Must be in degrees |
| Total for 7(c): 5 marks | ||
| Total for Question 7: 11 marks |
**7(a)**
| $A(0,1)$; $B(45°,0)$; $C(270°,-1)$ | B1, B1, B1, B1 | Condone radians; Condone (0.785,0) or better.; B1 for 270; B1 for −1 |
| Total for 7(a): 4 marks |
**7(b)**
| Stretch (I) in x-direction (II) with a scale factor $\frac{1}{2}$ (III) | M1A1 | More than one transformation is M0; M1 for (I) and either (II) or (III) |
| Total for 7(b): 2 marks |
**7(c)**
| $\cos^{-1}0.37 = "68.284..."$ ($=\alpha$); $x = \frac{\alpha}{2} = 34.1(42.)$; $x = 180 - \frac{\alpha}{2}$; $x = 180 + \frac{\alpha}{2}$ and $x = 180 + 180 - \frac{\alpha}{2}$; $2x = 68.284...;291.715...;$; $428.284...;651.715...$; $x = (34.1;)$; $145.9;214.1;325.9$ | M1, A1, m1, m1, A1 | Cos$^{-1}$(0.37) (PI eg 68.3 or 1.19); Condone 34.2°, 34° or 0.596 rads; OE eg $2x = 360 - \alpha$; OE Need both (OE for $2x =$ ) with no extras (quadrants) within the given interval; Dep. on all three method marks. Must be in degrees |
| Total for 7(c): 5 marks |
| **Total for Question 7: 11 marks** |
---7 The diagram shows the graph of $y = \cos 2 x$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{4a4d4dcd-4137-427d-834f-ac2fe83f8aeb-4_518_906_1098_552}
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of the points $A , B$ and $C$ marked on the diagram.
\item Describe the single geometrical transformation by which the curve with equation $y = \cos 2 x$ can be obtained from the curve with equation $y = \cos x$.
\item Solve the equation
$$\cos 2 x = 0.37$$
giving all solutions to the nearest $0.1 ^ { \circ }$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$. (No credit will be given for simply reading values from a graph.)\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q7 [11]}}