Edexcel P2 2024 January — Question 9 14 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2024
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeMultiple angle equations
DifficultyStandard +0.3 This is a straightforward multi-part question testing standard techniques. Part (i) requires converting sin x tan x to sin²x/cos x and solving a quadratic in cos x—routine manipulation. Part (ii) involves reading amplitude from a graph, finding where a transformed sine curve equals 2 (its midline), and locating a maximum—all standard textbook exercises requiring recall of transformations rather than problem-solving insight.
Spec1.05f Trigonometric function graphs: symmetries and periodicities1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

  1. In this question you must show detailed reasoning.
Solutions relying entirely on calculator technology are not acceptable.
  1. Solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation $$\sin x \tan x = 5$$ giving your answers to one decimal place.
  2. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{0e08d931-aa1c-48a8-8b39-47096f981950-26_643_736_721_660} \captionsetup{labelformat=empty} \caption{Figure 1}
    \end{figure} Figure 1 shows a sketch of part of the curve with equation $$y = A \sin \left( 2 \theta - \frac { 3 \pi } { 8 } \right) + 2$$ where \(A\) is a constant and \(\theta\) is measured in radians.
    The points \(P , Q\) and \(R\) lie on the curve and are shown in Figure 1.
    Given that the \(y\) coordinate of \(P\) is 7
    (a) state the value of \(A\),
    (b) find the exact coordinates of \(Q\),
    (c) find the value of \(\theta\) at \(R\), giving your answer to 3 significant figures.

Question 9:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
States or uses \(\tan x = \dfrac{\sin x}{\cos x}\)B1 May be seen as \(\sin x\tan x \Rightarrow \tan^2 x\cos x\)
\(\sin x\tan x=5 \Rightarrow \sin^2 x=5\cos x \Rightarrow 1-\cos^2 x=5\cos x\)M1A1 Uses \(\pm\sin^2 x\pm\cos^2 x=\pm1\) to set up quadratic in \(\cos x\) only
\(\cos^2 x+5\cos x-1=0 \Rightarrow \cos x=\dfrac{-5\pm\sqrt{29}}{2} \Rightarrow x=\) awrt \(78.9°, 281.1°\)M1dM1A1 Must be in degrees; no other values in range
Part (ii)(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(A=5\)B1 Check by the question
Part (ii)(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2\theta-\dfrac{3\pi}{8}=\dfrac{3\pi}{2} \Rightarrow \theta=\ldots\)M1 Must use correct order of operations; mixture of radians/degrees without recovery: M0
\(\theta=\dfrac{15\pi}{16}\)A1 Must be exact
\(y\) coordinate \(Q=-3\) (or \(2-\)"\(A\)")B1ft May be seen on diagram
Part (ii)(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Sets \(0=\text{"5"}\sin\!\left(2\theta-\dfrac{3\pi}{8}\right)+2 \Rightarrow \sin\!\left(2\theta-\dfrac{3\pi}{8}\right)=\pm\dfrac{2}{\text{"5"}}\)M1 May be implied; solutions with no working score 0
\(\sin\!\left(2\theta-\dfrac{3\pi}{8}\right)=\pm\dfrac{2}{5} \Rightarrow 2\theta-\dfrac{3\pi}{8}=\arcsin\!\left(\pm\dfrac{2}{5}\right)=\ldots\)dM1 Allow use of \(X\) for \(2\theta\) or \(2\theta-\dfrac{3\pi}{8}\)
One of \(\theta=0.38, 2.4, 3.5, 5.5, 6.7, 8.6, 9.8\ldots\)A1
\(\theta=\) awrt \(5.51\)A1
Question 9 (Trigonometry):
AnswerMarks
dM1Proceeds from \(\sin\left(2\theta - \frac{3\pi}{8}\right) = \pm\frac{2}{5}\) to \(2\theta - \frac{3\pi}{8} = \arcsin\left(\pm\frac{2}{5}\right) = \ldots\) which is one of the values below:
\(2\theta - \frac{3\pi}{8} = \arcsin\left(-\frac{2}{5}\right) = -0.41, 3.6, 5.9, 9.8, 12.2\)
\(2\theta - \frac{3\pi}{8} = \arcsin\left(\frac{2}{5}\right) = 0.41, 2.7, 6.7, 9.0, 13.0, 15.3\)
May be implied by \(2\theta = \arcsin\left(\pm\frac{2}{5}\right) + 1.17\ldots\) or allow the expression \(= \dfrac{\arcsin\left(\pm\frac{2}{5}\right) + \frac{3\pi}{8}}{2}\)
The sign slip is only condoned before they take arcsin(…)
AnswerMarks Guidance
A1Any one of the values in the table provided M1dM1 has been scored. Do not withhold this mark if other incorrect angles are seen.
\(\theta\)Radians (awrt): \(0.38, 2.4, 3.5, 5.5, 6.7, 8.6, 9.8, 11.8, 12.9\) Degrees (awrt): \(22, 136, 202, 316, 382, 496, 562, 676, 742\)
A1\(\theta = \) awrt \(5.51\) only (must be in radians). Can only be scored from correct working and all previous marks are scored.
# Question 9:

## Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\tan x = \dfrac{\sin x}{\cos x}$ | B1 | May be seen as $\sin x\tan x \Rightarrow \tan^2 x\cos x$ |
| $\sin x\tan x=5 \Rightarrow \sin^2 x=5\cos x \Rightarrow 1-\cos^2 x=5\cos x$ | M1A1 | Uses $\pm\sin^2 x\pm\cos^2 x=\pm1$ to set up quadratic in $\cos x$ only |
| $\cos^2 x+5\cos x-1=0 \Rightarrow \cos x=\dfrac{-5\pm\sqrt{29}}{2} \Rightarrow x=$ awrt $78.9°, 281.1°$ | M1dM1A1 | Must be in degrees; no other values in range |

## Part (ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A=5$ | B1 | Check by the question |

## Part (ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\theta-\dfrac{3\pi}{8}=\dfrac{3\pi}{2} \Rightarrow \theta=\ldots$ | M1 | Must use correct order of operations; mixture of radians/degrees without recovery: M0 |
| $\theta=\dfrac{15\pi}{16}$ | A1 | Must be exact |
| $y$ coordinate $Q=-3$ (or $2-$"$A$") | B1ft | May be seen on diagram |

## Part (ii)(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $0=\text{"5"}\sin\!\left(2\theta-\dfrac{3\pi}{8}\right)+2 \Rightarrow \sin\!\left(2\theta-\dfrac{3\pi}{8}\right)=\pm\dfrac{2}{\text{"5"}}$ | M1 | May be implied; solutions with no working score 0 |
| $\sin\!\left(2\theta-\dfrac{3\pi}{8}\right)=\pm\dfrac{2}{5} \Rightarrow 2\theta-\dfrac{3\pi}{8}=\arcsin\!\left(\pm\dfrac{2}{5}\right)=\ldots$ | dM1 | Allow use of $X$ for $2\theta$ or $2\theta-\dfrac{3\pi}{8}$ |
| One of $\theta=0.38, 2.4, 3.5, 5.5, 6.7, 8.6, 9.8\ldots$ | A1 | |
| $\theta=$ awrt $5.51$ | A1 | |

## Question 9 (Trigonometry):

**dM1** | Proceeds from $\sin\left(2\theta - \frac{3\pi}{8}\right) = \pm\frac{2}{5}$ to $2\theta - \frac{3\pi}{8} = \arcsin\left(\pm\frac{2}{5}\right) = \ldots$ which is one of the values below:

$2\theta - \frac{3\pi}{8} = \arcsin\left(-\frac{2}{5}\right) = -0.41, 3.6, 5.9, 9.8, 12.2$

$2\theta - \frac{3\pi}{8} = \arcsin\left(\frac{2}{5}\right) = 0.41, 2.7, 6.7, 9.0, 13.0, 15.3$

May be implied by $2\theta = \arcsin\left(\pm\frac{2}{5}\right) + 1.17\ldots$ or allow the expression $= \dfrac{\arcsin\left(\pm\frac{2}{5}\right) + \frac{3\pi}{8}}{2}$

The sign slip is only condoned before they take arcsin(…)

**A1** | Any one of the values in the table provided M1dM1 has been scored. Do not withhold this mark if other incorrect angles are seen.

| $\theta$ | Radians (awrt): $0.38, 2.4, 3.5, 5.5, 6.7, 8.6, 9.8, 11.8, 12.9$ | Degrees (awrt): $22, 136, 202, 316, 382, 496, 562, 676, 742$ |

**A1** | $\theta = $ awrt $5.51$ only (must be in radians). **Can only be scored from correct working and all previous marks are scored.**

---
\begin{enumerate}
  \item In this question you must show detailed reasoning.
\end{enumerate}

Solutions relying entirely on calculator technology are not acceptable.\\
(i) Solve, for $0 \leqslant x < 360 ^ { \circ }$, the equation

$$\sin x \tan x = 5$$

giving your answers to one decimal place.\\
(ii)

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0e08d931-aa1c-48a8-8b39-47096f981950-26_643_736_721_660}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of part of the curve with equation

$$y = A \sin \left( 2 \theta - \frac { 3 \pi } { 8 } \right) + 2$$

where $A$ is a constant and $\theta$ is measured in radians.\\
The points $P , Q$ and $R$ lie on the curve and are shown in Figure 1.\\
Given that the $y$ coordinate of $P$ is 7\\
(a) state the value of $A$,\\
(b) find the exact coordinates of $Q$,\\
(c) find the value of $\theta$ at $R$, giving your answer to 3 significant figures.

\begin{center}

\end{center}

\hfill \mbox{\textit{Edexcel P2 2024 Q9 [14]}}