| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Show then solve substituted equation |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard algebraic manipulation (converting tan to sin/cos, using Pythagorean identity) followed by solving a quadratic in cos θ and applying a simple linear transformation (θ → 3x). While it involves multiple steps, each technique is routine for AS-level students and requires no novel insight or complex problem-solving. |
| 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 |
|---|---|---|
| \(4\cos\theta - 1 = 2\sin\theta\tan\theta \Rightarrow 4\cos\theta - 1 = 2\sin\theta \times \frac{\sin\theta}{\cos\theta}\) | M1 | Recall and use \(\tan\theta = \frac{\sin\theta}{\cos\theta}\); cannot just be stated |
| \(\Rightarrow 4\cos^2\theta - \cos\theta = 2\sin^2\theta\) oe | A1 | Correct line with no fractional terms |
| \(\Rightarrow 4\cos^2\theta - \cos\theta = 2(1-\cos^2\theta)\) | M1 | Attempts to use \(1 - \cos^2\theta = \sin^2\theta\) |
| \(6\cos^2\theta - \cos\theta - 2 = 0\) * | A1* | Proceeds to correct answer with rigorous reasoning; no errors in notation or bracketing |
| Answer | Marks | Guidance |
|---|---|---|
| For attempting to solve given quadratic | M1 | Factorising: look for \((ay+b)(cy+d)\) where \(ac = \pm 6\), \(bd = \pm 2\); or quadratic formula |
| \(\cos 3x = \frac{2}{3}, -\frac{1}{2}\) | B1 | For the roots \(\frac{2}{3}, -\frac{1}{2}\) oe |
| \(x = \frac{1}{3}\arccos\left(\frac{2}{3}\right)\) or \(\frac{1}{3}\arccos\left(-\frac{1}{2}\right)\) | M1 | Finds at least one solution for \(x\) from \(\cos 3x\) within the given range |
| \(x = 40°, 80°\), awrt \(16.1°\) | A1 | These values only; withhold if any other values present; condone 40.0 and 80.0 |
## Question 12(a):
$4\cos\theta - 1 = 2\sin\theta\tan\theta \Rightarrow 4\cos\theta - 1 = 2\sin\theta \times \frac{\sin\theta}{\cos\theta}$ | M1 | Recall and use $\tan\theta = \frac{\sin\theta}{\cos\theta}$; **cannot just be stated**
$\Rightarrow 4\cos^2\theta - \cos\theta = 2\sin^2\theta$ oe | A1 | Correct line with no fractional terms
$\Rightarrow 4\cos^2\theta - \cos\theta = 2(1-\cos^2\theta)$ | M1 | Attempts to use $1 - \cos^2\theta = \sin^2\theta$
$6\cos^2\theta - \cos\theta - 2 = 0$ * | A1* | Proceeds to correct answer with rigorous reasoning; no errors in notation or bracketing
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## Question 12(b):
For attempting to solve given quadratic | M1 | Factorising: look for $(ay+b)(cy+d)$ where $ac = \pm 6$, $bd = \pm 2$; or quadratic formula
$\cos 3x = \frac{2}{3}, -\frac{1}{2}$ | B1 | For the roots $\frac{2}{3}, -\frac{1}{2}$ oe
$x = \frac{1}{3}\arccos\left(\frac{2}{3}\right)$ or $\frac{1}{3}\arccos\left(-\frac{1}{2}\right)$ | M1 | Finds at least one solution for $x$ from $\cos 3x$ within the given range
$x = 40°, 80°$, awrt $16.1°$ | A1 | These values **only**; withhold if any other values present; condone 40.0 and 80.0
---\begin{enumerate}
\item (a) Show that the equation
\end{enumerate}
$$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$
can be written in the form
$$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
(b) Hence solve, for $0 \leqslant x < 90 ^ { \circ }$
$$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$
giving your answers, where appropriate, to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)
\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q12 [8]}}