| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Solve shifted trig equation |
| Difficulty | Moderate -0.8 This is a multi-part question covering standard C2 trigonometry content: sketching tan x, using periodicity, basic manipulation to show tan θ = -1, applying a substitution to solve, and describing transformations. All parts are routine textbook exercises requiring recall and direct application of standard techniques with no problem-solving insight needed. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05f Trigonometric function graphs: symmetries and periodicities1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | M1, A1, A1 | |
| 7(b) | \(61°\); \(241°\) | B1, B1 |
| 7(c)(i) | \(\sin \theta = -\cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta} = -1\) | B1 |
| \(\Rightarrow \tan \theta = -1\) | ||
| 7(c)(ii) | \(\Rightarrow \tan(x - 20°) = -1\) | M1 |
| \(x - 20° = \tan^{-1}(-1)\) | ||
| \(x - 20° = 135°, 315° \ldots\) | m1 | |
| \(x = 155°; 335°\) | A1, A1ft | 4 marks |
| 7(d) | Translation \(\begin{pmatrix} 20 \\ 0 \end{pmatrix}\) | B1, B1 |
| 7(e) | \(f(x) = \tan 4x\) | B1 |
**7(a)** | | M1, A1, A1 | 3 marks | Correct shape of branch from $O$ {to $90°$} or correct shapes of branches from $90°$-$360°$; Complete graph for $0° \leq x \leq 360°$ (Asymptotes not explicitly required but graphs should show 'tendency'); Correct scaling on x-axis $0° \leq x \leq 360°$ |
**7(b)** | $61°$; $241°$ | B1, B1 | 2 marks | For $61°$; For $241°$ and no 'extras' in the interval $0° \leq x \leq 360°$ |
**7(c)(i)** | $\sin \theta = -\cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta} = -1$ | B1 | 1 mark | AG; be convinced that the identity $\frac{\sin \theta}{\cos \theta} = \tan \theta$ is known and validly used |
| | $\Rightarrow \tan \theta = -1$ | | |
**7(c)(ii)** | $\Rightarrow \tan(x - 20°) = -1$ | M1 | | |
| | $x - 20° = \tan^{-1}(-1)$ | | |
| | $x - 20° = 135°, 315° \ldots$ | m1 | |
| | $x = 155°; 335°$ | A1, A1ft | 4 marks | Ft on $(180 + \text{'155'})$ and no 'extras' in the given interval |
**7(d)** | Translation $\begin{pmatrix} 20 \\ 0 \end{pmatrix}$ | B1, B1 | 2 marks | 'Translation'/'translate(d)' Accept equivalent in words provided linked to 'translation/move/shift' (Note: B0B1 is possible) |
**7(e)** | $f(x) = \tan 4x$ | B1 | 1 mark | For $\tan 4x$ |
---7
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = \tan x$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\item Write down the two solutions of the equation $\tan x = \tan 61 ^ { \circ }$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $\sin \theta + \cos \theta = 0$, show that $\tan \theta = - 1$.
\item Hence solve the equation $\sin \left( x - 20 ^ { \circ } \right) + \cos \left( x - 20 ^ { \circ } \right) = 0$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\end{enumerate}\item Describe the single geometrical transformation that maps the graph of $y = \tan x$ onto the graph of $y = \tan \left( x - 20 ^ { \circ } \right)$.
\item The curve $y = \tan x$ is stretched in the $x$-direction with scale factor $\frac { 1 } { 4 }$ to give the curve with equation $y = \mathrm { f } ( x )$. Write down an expression for $\mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q7 [13]}}