| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Show equation reduces to tan form |
| Difficulty | Standard +0.3 Part (i) is a straightforward application of dividing to get tan(θ+30°) = 7/3, then solving—a routine technique. Part (ii)(a) requires factoring sin x and using sin²x = 1-cos²x, which is standard manipulation. Part (ii)(b) involves solving a quadratic in cos x. All steps are textbook techniques with no novel insight required, making this 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 | Mark | Guidance |
| \(3\sin(\theta+30°)=7\cos(\theta+30°) \Rightarrow \tan(\theta+30°)=\frac{7}{3}\) | B1 | Divides through by \(\cos(\theta+30°)\) to get correct equation in tan |
| \(\theta+30°=\arctan\left(\frac{7}{3}\right)\ (=66.8°)\) | M1 | Applies arctan to find a value for \(\theta+30°\); allow if radians used |
| \(\theta=36.8°\) or \(216.8°\) (awrt) | A1A1 | A1 one of awrt \(36.8°\) or awrt \(216.8°\); A1 both with no other solutions in range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin(\theta+30°)=\frac{7}{\sqrt{58}}\) or \(\cos(\theta+30°)=\frac{3}{\sqrt{58}}\) | B1 | |
| Squaring both sides, applying \(\sin^2(...)+\cos^2(...)=1\) to achieve equation in just sin or cos, then applying inverse function | M1 | Must reject extra solutions from squaring |
| \(\theta=36.8°\) or \(216.8°\) (awrt) | A1A1 | As main scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3\sqrt{3}}{2}\sin\theta+\frac{3}{2}\cos\theta=\frac{7\sqrt{3}}{2}\cos\theta-\frac{7}{2}\sin\theta\) | B1 | Correct application of compound angle formula and trig ratios |
| Rearranges to \(\frac{\sin\theta}{\cos\theta}=\tan\theta=...\Rightarrow\theta=\arctan(...)=...\) | M1 | Attempts compound angle formula, rearranges for \(\tan\theta\), applies arctan |
| \(\theta=36.8°\) or \(216.8°\) (awrt) | A1A1 | As main scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (a) \(3\sin^3x=5\sin x-7\sin x\cos x \Rightarrow 3\sin x(1-\cos^2x)=5\sin x-7\sin x\cos x\) | M1 | Uses \(\sin^2x=1-\cos^2x\), gathers terms one side, factors out \(\sin x\); allow if \(\sin x\) cancelled |
| \(\Rightarrow \sin x(3\cos^2x-7\cos x+2)=0\) | A1 | Correct form; accept non-zero multiples of quadratic in \(\cos x\) |
| (b) \(\sin x=0\Rightarrow x=0\) | B1 | \(x=0\) as one solution |
| \(\Rightarrow \sin x(3\cos x-1)(\cos x-2)=0\Rightarrow\cos x=...\) | M1 | Solves quadratic in \(\cos x\), usual rules |
| \(3\cos x=1\Rightarrow x=\arccos\left(\frac{1}{3}\right)=(\pm)1.23\) | M1 | Takes arccos of at least one value with modulus less than 1 which is a root of the quadratic |
| Both of \(x=\) awrt \(-1.23\), \(1.23\) | A1 | Both solutions from cosine equation correct, no others in range |
## Question 8(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\sin(\theta+30°)=7\cos(\theta+30°) \Rightarrow \tan(\theta+30°)=\frac{7}{3}$ | **B1** | Divides through by $\cos(\theta+30°)$ to get correct equation in tan |
| $\theta+30°=\arctan\left(\frac{7}{3}\right)\ (=66.8°)$ | **M1** | Applies arctan to find a value for $\theta+30°$; allow if radians used |
| $\theta=36.8°$ or $216.8°$ (awrt) | **A1A1** | A1 one of awrt $36.8°$ or awrt $216.8°$; A1 both with no other solutions in range |
**Alt 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin(\theta+30°)=\frac{7}{\sqrt{58}}$ or $\cos(\theta+30°)=\frac{3}{\sqrt{58}}$ | **B1** | |
| Squaring both sides, applying $\sin^2(...)+\cos^2(...)=1$ to achieve equation in just sin or cos, then applying inverse function | **M1** | Must reject extra solutions from squaring |
| $\theta=36.8°$ or $216.8°$ (awrt) | **A1A1** | As main scheme |
**Alt 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3\sqrt{3}}{2}\sin\theta+\frac{3}{2}\cos\theta=\frac{7\sqrt{3}}{2}\cos\theta-\frac{7}{2}\sin\theta$ | **B1** | Correct application of compound angle formula and trig ratios |
| Rearranges to $\frac{\sin\theta}{\cos\theta}=\tan\theta=...\Rightarrow\theta=\arctan(...)=...$ | **M1** | Attempts compound angle formula, rearranges for $\tan\theta$, applies arctan |
| $\theta=36.8°$ or $216.8°$ (awrt) | **A1A1** | As main scheme |
---
## Question 8(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(a)** $3\sin^3x=5\sin x-7\sin x\cos x \Rightarrow 3\sin x(1-\cos^2x)=5\sin x-7\sin x\cos x$ | **M1** | Uses $\sin^2x=1-\cos^2x$, gathers terms one side, factors out $\sin x$; allow if $\sin x$ cancelled |
| $\Rightarrow \sin x(3\cos^2x-7\cos x+2)=0$ | **A1** | Correct form; accept non-zero multiples of quadratic in $\cos x$ |
| **(b)** $\sin x=0\Rightarrow x=0$ | **B1** | $x=0$ as one solution |
| $\Rightarrow \sin x(3\cos x-1)(\cos x-2)=0\Rightarrow\cos x=...$ | **M1** | Solves quadratic in $\cos x$, usual rules |
| $3\cos x=1\Rightarrow x=\arccos\left(\frac{1}{3}\right)=(\pm)1.23$ | **M1** | Takes arccos of at least one value with modulus less than 1 which is a root of the quadratic |
| Both of $x=$ awrt $-1.23$, $1.23$ | **A1** | Both solutions from cosine equation correct, no others in range |
---8. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\begin{enumerate}[label=(\roman*)]
\item Solve, for $0 < \theta < 360 ^ { \circ }$, the equation
$$3 \sin \left( \theta + 30 ^ { \circ } \right) = 7 \cos \left( \theta + 30 ^ { \circ } \right)$$
giving your answers to one decimal place.
\item (a) Show that the equation
$$3 \sin ^ { 3 } x = 5 \sin x - 7 \sin x \cos x$$
can be written in the form
$$\sin x \left( a \cos ^ { 2 } x + b \cos x + c \right) = 0$$
where $a , b$ and $c$ are constants to be found.\\
(b) Hence solve for $- \frac { \pi } { 2 } \leqslant x \leqslant \frac { \pi } { 2 }$ the equation
$$3 \sin ^ { 3 } x = 5 \sin x - 7 \sin x \cos x$$
\includegraphics[max width=\textwidth, alt={}, center]{515f245f-9c5b-4263-ab2c-0a4f96f3bff0-27_2644_1840_118_111}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2021 Q8 [10]}}