| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a standard algebraic manipulation of trigonometric equations requiring systematic application of tan θ = sin θ/cos θ, clearing denominators, and factoring to solve a cubic. Part (b) is routine substitution with double angles. Slightly easier than average due to clear scaffolding and straightforward algebraic steps. |
| Spec | 1.05j 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 |
| \(\frac{3\sin\theta\cos\theta}{2\sin\theta-1}=5\tan\theta\Rightarrow\frac{3\sin\theta\cos\theta}{2\sin\theta-1}=5\frac{\sin\theta}{\cos\theta}\) | M1 | Attempts to use \(\tan\theta=\frac{\sin\theta}{\cos\theta}\); \(5\tan\theta=\frac{5\sin\theta}{5\cos\theta}\) is M0 unless correct identity stated first |
| \(\frac{3\sin\theta\cos\theta}{2\sin\theta-1}=\frac{5\sin\theta}{\cos\theta}\Rightarrow3\sin\theta\cos^2\theta=5\sin\theta(2\sin\theta-1)\) | M1 | Multiplies both sides by both denominators; condone invisible brackets |
| \(3\sin\theta\cos^2\theta=10\sin^2\theta-5\sin\theta\Rightarrow3\sin\theta(1-\sin^2\theta)=10\sin^2\theta-5\sin\theta\) * | M1 | Uses \(\cos^2\theta=1-\sin^2\theta\) to form equation in \(\sin\theta\) only |
| \(\Rightarrow3\sin^3\theta+10\sin^2\theta-8\sin\theta=0\) | A1* | Achieves given answer with no errors; consistent notation required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3\sin^32x+10\sin^22x-8\sin2x=0\Rightarrow\sin2x(3\sin^22x+10\sin2x-8)=0\) then \((3\sin2x-2)(\sin2x+4)=0\Rightarrow\sin2x=...\) | M1 | Takes out factor \(\sin2x\), solves quadratic; must use part (a) result |
| \(\sin2x=\frac{2}{3}\) | A1 | Or allow \(\sin\theta=\frac{2}{3}\); do not accept \(x=\frac{2}{3}\) unless recovered |
| \((x=)\ 0.365\) | A1 | awrt \(0.365\); no additional solutions in range other than \(0\); degrees answer is A0 |
| \(x=0\) | B1 | Must be \(x\); condone \(0°\) and \(0.000\) |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3\sin\theta\cos\theta}{2\sin\theta-1}=5\tan\theta\Rightarrow\frac{3\sin\theta\cos\theta}{2\sin\theta-1}=5\frac{\sin\theta}{\cos\theta}$ | M1 | Attempts to use $\tan\theta=\frac{\sin\theta}{\cos\theta}$; $5\tan\theta=\frac{5\sin\theta}{5\cos\theta}$ is M0 unless correct identity stated first |
| $\frac{3\sin\theta\cos\theta}{2\sin\theta-1}=\frac{5\sin\theta}{\cos\theta}\Rightarrow3\sin\theta\cos^2\theta=5\sin\theta(2\sin\theta-1)$ | M1 | Multiplies both sides by both denominators; condone invisible brackets |
| $3\sin\theta\cos^2\theta=10\sin^2\theta-5\sin\theta\Rightarrow3\sin\theta(1-\sin^2\theta)=10\sin^2\theta-5\sin\theta$ * | M1 | Uses $\cos^2\theta=1-\sin^2\theta$ to form equation in $\sin\theta$ only |
| $\Rightarrow3\sin^3\theta+10\sin^2\theta-8\sin\theta=0$ | A1* | Achieves given answer with no errors; consistent notation required |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\sin^32x+10\sin^22x-8\sin2x=0\Rightarrow\sin2x(3\sin^22x+10\sin2x-8)=0$ then $(3\sin2x-2)(\sin2x+4)=0\Rightarrow\sin2x=...$ | M1 | Takes out factor $\sin2x$, solves quadratic; must use part (a) result |
| $\sin2x=\frac{2}{3}$ | A1 | Or allow $\sin\theta=\frac{2}{3}$; do not accept $x=\frac{2}{3}$ unless recovered |
| $(x=)\ 0.365$ | A1 | awrt $0.365$; no additional solutions in range other than $0$; degrees answer is A0 |
| $x=0$ | B1 | Must be $x$; condone $0°$ and $0.000$ |\begin{enumerate}
\item (a) Show that the equation
\end{enumerate}
$$\frac { 3 \sin \theta \cos \theta } { 2 \sin \theta - 1 } = 5 \tan \theta \quad \sin \theta \neq \frac { 1 } { 2 }$$
can be written in the form
$$3 \sin ^ { 3 } \theta + 10 \sin ^ { 2 } \theta - 8 \sin \theta = 0$$
(b) Hence solve, for $- \frac { \pi } { 4 } < x < \frac { \pi } { 4 }$
$$\frac { 3 \sin 2 x \cos 2 x } { 2 \sin 2 x - 1 } = 5 \tan 2 x$$
giving your answers to 3 decimal places where appropriate.\\
\hfill \mbox{\textit{Edexcel P2 2021 Q6 [8]}}