| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Find maximum or minimum value |
| Difficulty | Challenging +1.2 Part (a) is a standard harmonic form conversion requiring routine application of R cos(θ-α) = R cos α cos θ + R sin α sin θ, solving for R and α. Part (b) requires finding S_9 for an arithmetic sequence (standard formula), then expressing it in harmonic form to find the maximum—this involves multiple steps but uses well-practiced techniques. The connection between parts and the deduction in (b)(ii) adds modest problem-solving demand beyond pure recall, placing it slightly above average difficulty. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = \sqrt{2^2+8^2} = \sqrt{68} = 2\sqrt{17}\) | B1 | Accept \(2\sqrt{17}\) or \(\sqrt{68}\); \(\pm\) values score B0 |
| \(2\cos\theta + 8\sin\theta = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha\), so \(2 = R\cos\alpha\), \(8 = R\sin\alpha\), \(\tan\alpha = \frac{8}{2} \Rightarrow \alpha = \ldots\) | M1 | Proceeds to value for \(\alpha\) from \(\tan\alpha = \pm\frac{8}{2}\), \(\cos\alpha = \pm\frac{2}{\sqrt{68}}\), or \(\sin\alpha = \pm\frac{8}{\sqrt{68}}\) |
| \(\alpha =\) awrt \(1.326\) | A1 | awrt 1.326 for \(\alpha\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4.5 \times 2\sqrt{17}\) | M1 | For \(\pm4.5\times\) their \(R\); not embedded in expression unless extracted |
| \(9\sqrt{17}\) | A1 | Accept exact equivalents e.g. \(\sqrt{1377}\), \(4.5\sqrt{68}\), \(4.5(2\sqrt{17})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| awrt \(1.33\) | B1ft | Follow through on their \(\alpha\) even if in degrees (76°) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = \sqrt{2^2+8^2} = \sqrt{68} = 2\sqrt{17}$ | B1 | Accept $2\sqrt{17}$ or $\sqrt{68}$; $\pm$ values score B0 |
| $2\cos\theta + 8\sin\theta = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$, so $2 = R\cos\alpha$, $8 = R\sin\alpha$, $\tan\alpha = \frac{8}{2} \Rightarrow \alpha = \ldots$ | M1 | Proceeds to value for $\alpha$ from $\tan\alpha = \pm\frac{8}{2}$, $\cos\alpha = \pm\frac{2}{\sqrt{68}}$, or $\sin\alpha = \pm\frac{8}{\sqrt{68}}$ |
| $\alpha =$ awrt $1.326$ | A1 | awrt 1.326 for $\alpha$ |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4.5 \times 2\sqrt{17}$ | M1 | For $\pm4.5\times$ their $R$; not embedded in expression unless extracted |
| $9\sqrt{17}$ | A1 | Accept exact equivalents e.g. $\sqrt{1377}$, $4.5\sqrt{68}$, $4.5(2\sqrt{17})$ |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| awrt $1.33$ | B1ft | Follow through on their $\alpha$ even if in degrees (76°) |
---\begin{enumerate}
\item (a) Express $2 \cos \theta + 8 \sin \theta$ in the form $R \cos ( \theta - \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$\\
Give the exact value of $R$ and give the value of $\alpha$ in radians to 3 decimal places.
\end{enumerate}
The first three terms of an arithmetic sequence are
$$\cos x \quad \cos x + \sin x \quad \cos x + 2 \sin x \quad x \neq n \pi$$
Given that $S _ { 9 }$ represents the sum of the first 9 terms of this sequence as $x$ varies,\\
(b) (i) find the exact maximum value of $S _ { 9 }$\\
(ii) deduce the smallest positive value of $x$ at which this maximum value of $S _ { 9 }$ occurs.
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q8 [6]}}