| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 Part (a) is a routine algebraic manipulation using standard identities (tan θ = sin θ/cos θ and sin²θ + cos²θ = 1) to convert to quadratic form—straightforward bookwork. Part (b) applies the result with a double angle substitution and requires solving a quadratic, then finding angles in a given range. This is slightly above pure recall but remains a standard multi-step question with well-signposted techniques and no novel insight required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta(3\cos\theta - \frac{\sin\theta}{\cos\theta}) = 2\) giving \(3\cos^2\theta - \sin\theta = 2\) | M1 | Uses \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) to write in terms of sine and cosine |
| \(3(1-\sin^2\theta) - \sin\theta = 2\) | M1 | Uses \(\sin^2\theta + \cos^2\theta = 1\) to get quadratic in sine only |
| \(3\sin^2\theta + \sin\theta - 1 = 0\) * | A1* | Condone notational slip e.g. \(3\sin\theta^2\); printed answer must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin 2x = \frac{-1 \pm \sqrt{13}}{6}\) (or awrt \(0.43\) and awrt \(-0.77\)) | M1A1 | Solving quadratic \(3\sin^2\theta + \sin\theta - 1 = 0\) |
| \(2x = \sin^{-1}(0.434...)\) or \(2x = \sin^{-1}(-0.77...)\) | M1 | Attempts inverse sine and divides by 2 |
| \(-1.13,\ -0.438,\ 0.225,\ 1.35\) | A1A1 | First A1: any two correct; Second A1: all four correct, no extras in range |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta(3\cos\theta - \frac{\sin\theta}{\cos\theta}) = 2$ giving $3\cos^2\theta - \sin\theta = 2$ | M1 | Uses $\tan\theta = \frac{\sin\theta}{\cos\theta}$ to write in terms of sine and cosine |
| $3(1-\sin^2\theta) - \sin\theta = 2$ | M1 | Uses $\sin^2\theta + \cos^2\theta = 1$ to get quadratic in sine only |
| $3\sin^2\theta + \sin\theta - 1 = 0$ * | A1* | Condone notational slip e.g. $3\sin\theta^2$; printed answer must be correct |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin 2x = \frac{-1 \pm \sqrt{13}}{6}$ (or awrt $0.43$ and awrt $-0.77$) | M1A1 | Solving quadratic $3\sin^2\theta + \sin\theta - 1 = 0$ |
| $2x = \sin^{-1}(0.434...)$ or $2x = \sin^{-1}(-0.77...)$ | M1 | Attempts inverse sine and divides by 2 |
| $-1.13,\ -0.438,\ 0.225,\ 1.35$ | A1A1 | First A1: any two correct; Second A1: all four correct, no extras in range |
Special case in degrees: $-64.9, -25.1, 12.9, 77.1$ — score A1 for any two, then A0 (max 7/8)
---\begin{enumerate}
\item In this question you must show detailed reasoning.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Show that the equation
$$( 3 \cos \theta - \tan \theta ) \cos \theta = 2$$
can be written as
$$3 \sin ^ { 2 } \theta + \sin \theta - 1 = 0$$
(b) Hence solve for $- \frac { \pi } { 2 } \leqslant x \leqslant \frac { \pi } { 2 }$
$$( 3 \cos 2 x - \tan 2 x ) \cos 2 x = 2$$
\hfill \mbox{\textit{Edexcel P2 2022 Q5 [8]}}