| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a structured algebraic manipulation question with clear guidance. Part (a) requires systematic use of standard identities (tan θ = sin θ/cos θ, sin 2θ = 2sin θ cos θ, 1 + tan²θ = sec²θ) to reach a given form—methodical but not conceptually demanding. Part (b) is routine solving once the quadratic form is established. The question scaffolds the approach and uses familiar Further Maths Pure content, making it slightly easier than average overall. |
| 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 |
| Uses at least one correct trig identity e.g. \(2\frac{\sin\theta}{\cos\theta}(8\cos\theta + 23(1-\cos^2\theta)) = 8\times 2\sin\theta\cos\theta\sec^2\theta\) | B1 | Recalling and using at least one correct trig identity |
| \(\Rightarrow 2\sin\theta\cos\theta(8\cos\theta + 23(1-\cos^2\theta)) = 8\sin 2\theta\) | M1 | AO 2.1 — manipulating using trig identities (condone sign slips only) to obtain \(A\sin 2\theta\cos^2\theta + B\sin 2\theta\cos\theta + C\sin 2\theta\ (=0)\) |
| \(\sin 2\theta(8\cos\theta + 23(1-\cos^2\theta)) = 8\sin 2\theta\) | AO 2.2a | |
| \(\sin 2\theta(23\cos^2\theta - 8\cos\theta - 15) = 0\) | M1A1 | AO 2.2a — note not a given answer so condone notational slips; "= 0" required for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sin 2x(23\cos^2 x - 8\cos x - 15) = 0\) | ||
| \(\sin 2x = 0 \Rightarrow x = 360°\) or \(540°\) | B1 | AO 2.2a — condone \(x=2\pi\) or \(x=3\pi\); degrees symbol not required |
| \(23\cos^2 x - 8\cos x - 15 \Rightarrow \cos x = -\frac{15}{23}\) | M1 | AO 1.1b — attempts to solve 3TQ to a value for \(\cos x\) |
| \(\cos x = -\frac{15}{23} \Rightarrow x = \ldots\) | dM1 | AO 1.1b — attempts to find angle in range \(360° < x < 540°\) (not 450) for \(\cos x = k\), \( |
| \(x = 360°,\ 540°\) and awrt \(491°\) only | A1 | AO 2.3 — no other values in range (including 450) |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses at least one correct trig identity e.g. $2\frac{\sin\theta}{\cos\theta}(8\cos\theta + 23(1-\cos^2\theta)) = 8\times 2\sin\theta\cos\theta\sec^2\theta$ | B1 | Recalling and using at least one correct trig identity |
| $\Rightarrow 2\sin\theta\cos\theta(8\cos\theta + 23(1-\cos^2\theta)) = 8\sin 2\theta$ | M1 | AO 2.1 — manipulating using trig identities (condone sign slips only) to obtain $A\sin 2\theta\cos^2\theta + B\sin 2\theta\cos\theta + C\sin 2\theta\ (=0)$ |
| $\sin 2\theta(8\cos\theta + 23(1-\cos^2\theta)) = 8\sin 2\theta$ | | AO 2.2a |
| $\sin 2\theta(23\cos^2\theta - 8\cos\theta - 15) = 0$ | M1A1 | AO 2.2a — note not a given answer so condone notational slips; "= 0" required for A1 |
**(3 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin 2x(23\cos^2 x - 8\cos x - 15) = 0$ | | |
| $\sin 2x = 0 \Rightarrow x = 360°$ **or** $540°$ | B1 | AO 2.2a — condone $x=2\pi$ or $x=3\pi$; degrees symbol not required |
| $23\cos^2 x - 8\cos x - 15 \Rightarrow \cos x = -\frac{15}{23}$ | M1 | AO 1.1b — attempts to solve 3TQ to a value for $\cos x$ |
| $\cos x = -\frac{15}{23} \Rightarrow x = \ldots$ | dM1 | AO 1.1b — attempts to find angle in range $360° < x < 540°$ (not 450) for $\cos x = k$, $|k|<1$ |
| $x = 360°,\ 540°$ and awrt $491°$ only | A1 | AO 2.3 — no other values in range (including 450) |
**(4 marks) — (7 marks total)**
---\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
$$2 \tan \theta \left( 8 \cos \theta + 23 \sin ^ { 2 } \theta \right) = 8 \sin 2 \theta \left( 1 + \tan ^ { 2 } \theta \right)$$
may be written as
$$\sin 2 \theta \left( A \cos ^ { 2 } \theta + B \cos \theta + C \right) = 0$$
where $A , B$ and $C$ are constants to be found.\\
(b) Hence, solve for $360 ^ { \circ } \leqslant x \leqslant 540 ^ { \circ }$
$$2 \tan x \left( 8 \cos x + 23 \sin ^ { 2 } x \right) = 8 \sin 2 x \left( 1 + \tan ^ { 2 } x \right) \quad x \in \mathbb { R } \quad x \neq 450 ^ { \circ }$$
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q14 [7]}}