Edexcel P3 2024 January — Question 9 8 marks

Exam BoardEdexcel
ModuleP3 (Pure Mathematics 3)
Year2024
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeShow equation reduces to tan form
DifficultyStandard +0.3 This is a standard P3 double angle formula question requiring algebraic manipulation (expanding, using sin 2θ = 2sin θ cos θ, cos 2θ = cos²θ - sin²θ, and sec 2θ = 1/cos 2θ) followed by routine solving of R-form trigonometric equation. The steps are methodical and follow well-practiced techniques with no novel insight required, making it slightly easier than average.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals1.05p Proof involving trig: functions and identities

  1. In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
  1. Show that the equation $$\frac { 3 \sin \theta \cos \theta } { \cos \theta + \sin \theta } = ( 2 + \sec 2 \theta ) ( \cos \theta - \sin \theta )$$ can be written in the form $$3 \sin 2 \theta - 4 \cos 2 \theta = 2$$
  2. Hence solve for \(\pi < x < \frac { 3 \pi } { 2 }\) $$\frac { 3 \sin x \cos x } { \cos x + \sin x } = ( 2 + \sec 2 x ) ( \cos x - \sin x )$$ giving the answer to 3 significant figures.

Question 9:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{3\sin\theta\cos\theta}{\cos\theta+\sin\theta} = (2+\sec 2\theta)(\cos\theta-\sin\theta)\) \(\Rightarrow \frac{3}{2}\sin 2\theta = (2+\sec 2\theta)(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)\)M1 Attempts to use \(\sin 2\theta = 2\sin\theta\cos\theta\) or \(\cos 2\theta = \pm\cos^2\theta \pm \sin^2\theta\). Accept going from \((\cos\theta-\sin\theta)(\cos\theta+\sin\theta)\) to \(\cos 2\theta\)
\(\Rightarrow \frac{3}{2}\sin 2\theta = (2+\sec 2\theta)\cos 2\theta\)dM1 Attempts to use \(\sin 2\theta = 2\sin\theta\cos\theta\) and \(\cos 2\theta = \pm\cos^2\theta \pm \sin^2\theta\). Depends on previous M mark.
\(\Rightarrow \frac{3}{2}\sin 2\theta = 2\cos 2\theta + 1 \Rightarrow 3\sin 2\theta - 4\cos 2\theta = 2\ *\)A1* Uses \(\sec 2\theta = \frac{1}{\cos 2\theta}\) to achieve equation in just \(\sin 2\theta\) and \(\cos 2\theta\); all previous working shown. Note: accept \(6\sin\theta\cos\theta = 4\cos 2\theta + 2\) to given answer. Condone missing \(\theta\) or missing bracket but withhold A mark for clear errors such as missing "2" or missing pair of brackets not recovered.
Question (b):
Way 1 – Using \(R\sin(2x - \alpha)\) or \(R\cos(2x + \alpha)\)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(R = \sqrt{3^2 + 4^2} = 5\) or \(\alpha = \tan^{-1}\left(\frac{4}{3} \text{ or } \frac{3}{4}\right)\)M1 Condone obtaining \(\pm 5\); or \(\alpha = \tan^{-1}\left(\pm\frac{4}{3}\right)\), \(\tan^{-1}\left(\pm\frac{3}{4}\right)\), \(\cos^{-1}\left(\pm\frac{3}{5}\right)\), \(\sin^{-1}\left(\pm\frac{4}{5}\right)\), \(\cos^{-1}\left(\pm\frac{4}{5}\right)\), \(\sin^{-1}\left(\pm\frac{3}{5}\right)\)
\(R = \sqrt{3^2 + 4^2} = 5\) and \(\alpha = \tan^{-1}\left(\frac{4}{3} \text{ or } \frac{3}{4}\right)\)dM1 Both required; depends on previous M1
\(5\sin(2x - 0.927) = 2\) or \(5\cos(2x + 0.644) = -2\)A1 Accept awrt 0.93 or 0.64 for \(\alpha\); do not allow \(\pm 5\sin(2x-0.927)=2\) unless recovered
\(x = \dfrac{\sin^{-1}\frac{2}{5} + 0.927}{2}\) or \(x = \dfrac{\cos^{-1}\frac{-2}{5} - 0.644}{2}\)ddM1 Correct order of operations; condone \(\pm\)"0.927" or \(\pm\)"0.644"; depends on both previous method marks
\(x = \text{awrt } 3.81\)A1 No other values in range; \(x = \pi + 0.669\) scores A0
Way 2 – Squaring
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(3\sin 2x = 2 + 4\cos 2x \Rightarrow 9\sin^2 2x = 4 + 16\cos 2x + 16\cos^2 2x\)M1 Must isolate one trig term then square both sides correctly; do not condone poor squaring
\(25\cos^2 2x + 16\cos 2x - 5 = 0\) or \(25\sin^2 2x - 12\sin 2x - 12 = 0\)dM1, A1 Uses \(\sin^2 2x + \cos^2 2x = 1\); proceeds to 3-term quadratic in \(\sin 2x\) or \(\cos 2x\)
\(\cos 2x = \dfrac{-8 \pm 3\sqrt{21}}{25}\) (0.23, \(-\)0.87) or \(\sin 2x = \dfrac{6 \pm 4\sqrt{21}}{25}\) (0.97, \(-\)0.49)ddM1 Attempts to solve quadratic; depends on both previous method marks
\(x = \text{awrt } 3.81\)A1 No other values in range
Way 3 – Using double angle formulae
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(6\sin x\cos x - 4\cos^2 x + 4\sin^2 x = 2\)M1 Uses \(\sin 2x = 2\sin x\cos x\) and \(\cos 2x = \pm\sin^2 x \pm \cos^2 x\)
\(\tan^2 x + 3\tan x - 3 = 0\)dM1, A1 Divides by \(\cos^2 x\), uses \(\sec^2 x = 1 + \tan^2 x\); 3-term quadratic in \(\tan x\)
\(\tan x = \dfrac{-3 \pm \sqrt{21}}{2}\) (0.79, \(-\)3.79)ddM1 Solves quadratic; depends on both previous method marks
\(x = \text{awrt } 3.81\)A1 No other values in range
Way 4 – Divides by \(\cos 2x\) and squares
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(3\tan 2x - 4 = 2\sec 2x \Rightarrow 9\tan^2 2x - 24\tan 2x + 16 = 4\sec^2 2x\)M1 Divides by \(\cos 2x\) then squares; do not condone poor squaring e.g. \((3\tan 2x-4)^2 = 9\tan^2 2x+16\)
\(5\tan^2 2x - 24\tan 2x + 12 = 0\)dM1, A1 Uses \(\sec^2 2x = 1 + \tan^2 2x\); 3-term quadratic in \(\tan 2x\)
\(\tan 2x = \dfrac{12 \pm 2\sqrt{21}}{5}\) (4.23, 0.57)ddM1 Solves quadratic; depends on both previous method marks
\(x = \text{awrt } 3.81\)A1 No other values in range; \(x = \pi + 0.669\) scores A0
Total: 5 marks for part (b)
## Question 9:

**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3\sin\theta\cos\theta}{\cos\theta+\sin\theta} = (2+\sec 2\theta)(\cos\theta-\sin\theta)$ $\Rightarrow \frac{3}{2}\sin 2\theta = (2+\sec 2\theta)(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)$ | M1 | Attempts to use $\sin 2\theta = 2\sin\theta\cos\theta$ **or** $\cos 2\theta = \pm\cos^2\theta \pm \sin^2\theta$. Accept going from $(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)$ to $\cos 2\theta$ |
| $\Rightarrow \frac{3}{2}\sin 2\theta = (2+\sec 2\theta)\cos 2\theta$ | dM1 | Attempts to use $\sin 2\theta = 2\sin\theta\cos\theta$ **and** $\cos 2\theta = \pm\cos^2\theta \pm \sin^2\theta$. Depends on previous M mark. |
| $\Rightarrow \frac{3}{2}\sin 2\theta = 2\cos 2\theta + 1 \Rightarrow 3\sin 2\theta - 4\cos 2\theta = 2\ *$ | A1* | Uses $\sec 2\theta = \frac{1}{\cos 2\theta}$ to achieve equation in just $\sin 2\theta$ and $\cos 2\theta$; all previous working shown. Note: accept $6\sin\theta\cos\theta = 4\cos 2\theta + 2$ to given answer. Condone missing $\theta$ or missing bracket but withhold A mark for clear errors such as missing "2" or missing pair of brackets not recovered. |

## Question (b):

**Way 1 – Using $R\sin(2x - \alpha)$ or $R\cos(2x + \alpha)$**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{3^2 + 4^2} = 5$ **or** $\alpha = \tan^{-1}\left(\frac{4}{3} \text{ or } \frac{3}{4}\right)$ | M1 | Condone obtaining $\pm 5$; or $\alpha = \tan^{-1}\left(\pm\frac{4}{3}\right)$, $\tan^{-1}\left(\pm\frac{3}{4}\right)$, $\cos^{-1}\left(\pm\frac{3}{5}\right)$, $\sin^{-1}\left(\pm\frac{4}{5}\right)$, $\cos^{-1}\left(\pm\frac{4}{5}\right)$, $\sin^{-1}\left(\pm\frac{3}{5}\right)$ |
| $R = \sqrt{3^2 + 4^2} = 5$ **and** $\alpha = \tan^{-1}\left(\frac{4}{3} \text{ or } \frac{3}{4}\right)$ | dM1 | Both required; depends on previous M1 |
| $5\sin(2x - 0.927) = 2$ **or** $5\cos(2x + 0.644) = -2$ | A1 | Accept awrt 0.93 or 0.64 for $\alpha$; do not allow $\pm 5\sin(2x-0.927)=2$ unless recovered |
| $x = \dfrac{\sin^{-1}\frac{2}{5} + 0.927}{2}$ **or** $x = \dfrac{\cos^{-1}\frac{-2}{5} - 0.644}{2}$ | ddM1 | Correct order of operations; condone $\pm$"0.927" or $\pm$"0.644"; depends on both previous method marks |
| $x = \text{awrt } 3.81$ | A1 | No other values in range; $x = \pi + 0.669$ scores A0 |

---

**Way 2 – Squaring**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\sin 2x = 2 + 4\cos 2x \Rightarrow 9\sin^2 2x = 4 + 16\cos 2x + 16\cos^2 2x$ | M1 | Must isolate one trig term then square both sides correctly; do not condone poor squaring |
| $25\cos^2 2x + 16\cos 2x - 5 = 0$ **or** $25\sin^2 2x - 12\sin 2x - 12 = 0$ | dM1, A1 | Uses $\sin^2 2x + \cos^2 2x = 1$; proceeds to 3-term quadratic in $\sin 2x$ or $\cos 2x$ |
| $\cos 2x = \dfrac{-8 \pm 3\sqrt{21}}{25}$ (0.23, $-$0.87) **or** $\sin 2x = \dfrac{6 \pm 4\sqrt{21}}{25}$ (0.97, $-$0.49) | ddM1 | Attempts to solve quadratic; depends on both previous method marks |
| $x = \text{awrt } 3.81$ | A1 | No other values in range |

---

**Way 3 – Using double angle formulae**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $6\sin x\cos x - 4\cos^2 x + 4\sin^2 x = 2$ | M1 | Uses $\sin 2x = 2\sin x\cos x$ and $\cos 2x = \pm\sin^2 x \pm \cos^2 x$ |
| $\tan^2 x + 3\tan x - 3 = 0$ | dM1, A1 | Divides by $\cos^2 x$, uses $\sec^2 x = 1 + \tan^2 x$; 3-term quadratic in $\tan x$ |
| $\tan x = \dfrac{-3 \pm \sqrt{21}}{2}$ (0.79, $-$3.79) | ddM1 | Solves quadratic; depends on both previous method marks |
| $x = \text{awrt } 3.81$ | A1 | No other values in range |

---

**Way 4 – Divides by $\cos 2x$ and squares**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\tan 2x - 4 = 2\sec 2x \Rightarrow 9\tan^2 2x - 24\tan 2x + 16 = 4\sec^2 2x$ | M1 | Divides by $\cos 2x$ then squares; do not condone poor squaring e.g. $(3\tan 2x-4)^2 = 9\tan^2 2x+16$ |
| $5\tan^2 2x - 24\tan 2x + 12 = 0$ | dM1, A1 | Uses $\sec^2 2x = 1 + \tan^2 2x$; 3-term quadratic in $\tan 2x$ |
| $\tan 2x = \dfrac{12 \pm 2\sqrt{21}}{5}$ (4.23, 0.57) | ddM1 | Solves quadratic; depends on both previous method marks |
| $x = \text{awrt } 3.81$ | A1 | No other values in range; $x = \pi + 0.669$ scores A0 |

**Total: 5 marks for part (b)**
\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

Solutions relying entirely on calculator technology are not acceptable.\\
(a) Show that the equation

$$\frac { 3 \sin \theta \cos \theta } { \cos \theta + \sin \theta } = ( 2 + \sec 2 \theta ) ( \cos \theta - \sin \theta )$$

can be written in the form

$$3 \sin 2 \theta - 4 \cos 2 \theta = 2$$

(b) Hence solve for $\pi < x < \frac { 3 \pi } { 2 }$

$$\frac { 3 \sin x \cos x } { \cos x + \sin x } = ( 2 + \sec 2 x ) ( \cos x - \sin x )$$

giving the answer to 3 significant figures.

\hfill \mbox{\textit{Edexcel P3 2024 Q9 [8]}}