| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Prove identity then solve equation |
| Difficulty | Standard +0.3 Part (a) is a standard bookwork proof using cos(2A+A) and double angle formulae that appears in most textbooks. Part (b) requires substituting the identity, using sin²x = 1-cos²x, then solving a cubic in cos x, which is routine for this topic. The question tests standard techniques without requiring novel insight, making it 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 |
|---|---|---|
| Working | Mark | Guidance |
| \(\cos 3A = \cos(2A+A) = \cos 2A\cos A - \sin 2A\sin A\) | M1 | 3.1a - Attempts compound angle formula for \(\cos(2A+A)\) or \(\cos(A+2A)\) |
| \(=(2\cos^2 A-1)\cos A-(2\sin A\cos A)\sin A\) | dM1 | 1.1b - Uses correct double angle identities for \(\cos 2A\) and \(\sin 2A\) |
| \(=(2\cos^2 A-1)\cos A - 2\cos A(1-\cos^2 A)\) | ddM1 | 2.1 - Attempts to get all terms in terms of \(\cos A\) |
| \(=4\cos^3 A - 3\cos A\) * | A1* | 1.1b - Completely correct and rigorous proof |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(1-\cos 3x = \sin^2 x \Rightarrow \cos^2 x + 3\cos x - 4\cos^3 x = 0\) | M1 | 1.1b - Produces equation in \(\cos x\) using part (a) and \(\sin^2 x = 1-\cos^2 x\) |
| \(\Rightarrow \cos x(4\cos^2 x - \cos x - 3)=0\) \(\Rightarrow \cos x(4\cos x+3)(\cos x-1)=0\) \(\Rightarrow \cos x = ...\) | dM1 | 3.1a - Takes cubic and makes valid attempt to solve |
| Two of \(-90°, 0, 90°\), awrt \(139°\) | A1 | 1.1b - Depends on first M mark |
| All four of \(-90°, 0, 90°\), awrt \(139°\) | A1 | 2.1 - No extra solutions in range |
# Question 10:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\cos 3A = \cos(2A+A) = \cos 2A\cos A - \sin 2A\sin A$ | M1 | 3.1a - Attempts compound angle formula for $\cos(2A+A)$ or $\cos(A+2A)$ |
| $=(2\cos^2 A-1)\cos A-(2\sin A\cos A)\sin A$ | dM1 | 1.1b - Uses correct double angle identities for $\cos 2A$ and $\sin 2A$ |
| $=(2\cos^2 A-1)\cos A - 2\cos A(1-\cos^2 A)$ | ddM1 | 2.1 - Attempts to get all terms in terms of $\cos A$ |
| $=4\cos^3 A - 3\cos A$ * | A1* | 1.1b - Completely correct and rigorous proof |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $1-\cos 3x = \sin^2 x \Rightarrow \cos^2 x + 3\cos x - 4\cos^3 x = 0$ | M1 | 1.1b - Produces equation in $\cos x$ using part (a) and $\sin^2 x = 1-\cos^2 x$ |
| $\Rightarrow \cos x(4\cos^2 x - \cos x - 3)=0$ $\Rightarrow \cos x(4\cos x+3)(\cos x-1)=0$ $\Rightarrow \cos x = ...$ | dM1 | 3.1a - Takes cubic and makes valid attempt to solve |
| Two of $-90°, 0, 90°$, awrt $139°$ | A1 | 1.1b - Depends on first M mark |
| All four of $-90°, 0, 90°$, awrt $139°$ | A1 | 2.1 - No extra solutions in range |
---\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
$$\cos 3 A \equiv 4 \cos ^ { 3 } A - 3 \cos A$$
(b) Hence solve, for $- 90 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$, the equation
$$1 - \cos 3 x = \sin ^ { 2 } x$$
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q10 [8]}}