| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Factorization method |
| Difficulty | Moderate -0.3 This is a straightforward C2 trigonometric equation requiring standard manipulation (converting tan to sin/cos) and solving a factored form. Part (a) is shown algebraically, part (b) requires solving sin 2x = 0 and cos 2x = 1/5 within a given range—routine techniques with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| States or uses \(\tan 2x = \frac{\sin 2x}{\cos 2x}\) | M1 | Statement that \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) or replacement of tan wherever it appears. Must be correct but may involve \(\theta\) instead of \(2x\) |
| \(\frac{\sin 2x}{\cos 2x} = 5\sin 2x \Rightarrow \sin 2x - 5\sin 2x\cos 2x = 0 \Rightarrow \sin 2x(1 - 5\cos 2x) = 0\) | A1 | All steps must be shown. \(\sin 2x = 5\sin 2x\cos 2x\) is not sufficient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin 2x = 0\) gives \(2x = 0, 180, 360\) so \(x = 0, 90, 180\) | B1, B1 | First B1 for two correct answers, second B1 for all three correct. Excess in range loses last B1 |
| \(\cos 2x = \frac{1}{5}\) gives \(2x = 78.46\) (or 78.5 or 78.4) or \(2x = 281.54\) (or 281.6) | M1 | Must relate to \(2x\) not just \(x\) |
| \(x = 39.2\) (or 39.3), \(140.8\) (or 141) | A1, A1 | First A1 for 39.2, second for 140.8 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| States or uses $\tan 2x = \frac{\sin 2x}{\cos 2x}$ | M1 | Statement that $\tan\theta = \frac{\sin\theta}{\cos\theta}$ or replacement of tan wherever it appears. Must be correct but may involve $\theta$ instead of $2x$ |
| $\frac{\sin 2x}{\cos 2x} = 5\sin 2x \Rightarrow \sin 2x - 5\sin 2x\cos 2x = 0 \Rightarrow \sin 2x(1 - 5\cos 2x) = 0$ | A1 | All steps must be shown. $\sin 2x = 5\sin 2x\cos 2x$ is not sufficient |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin 2x = 0$ gives $2x = 0, 180, 360$ so $x = 0, 90, 180$ | B1, B1 | First B1 for two correct answers, second B1 for all three correct. Excess in range loses last B1 |
| $\cos 2x = \frac{1}{5}$ gives $2x = 78.46$ (or 78.5 or 78.4) **or** $2x = 281.54$ (or 281.6) | M1 | Must relate to $2x$ not just $x$ |
| $x = 39.2$ (or 39.3), $140.8$ (or 141) | A1, A1 | First A1 for 39.2, second for 140.8 |
---\begin{enumerate}
\item (a) Show that the equation
\end{enumerate}
$$\tan 2 x = 5 \sin 2 x$$
can be written in the form
$$( 1 - 5 \cos 2 x ) \sin 2 x = 0$$
(b) Hence solve, for $0 \leqslant x \leqslant 180 ^ { \circ }$,
$$\tan 2 x = 5 \sin 2 x$$
giving your answers to 1 decimal place where appropriate.\\
You must show clearly how you obtained your answers.\\
\hfill \mbox{\textit{Edexcel C2 2012 Q6 [7]}}