| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Find value where max/min occurs |
| Difficulty | Moderate -0.3 This is a standard harmonic form question requiring the routine application of R sin(x - α) = R sin x cos α - R cos x sin α, comparing coefficients to find R and α, then using the known maximum value of sine. While it requires multiple steps, the technique is algorithmic and commonly practiced in C4, making it slightly easier than average. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin x - \sqrt{3}\cos x = R\sin(x-\alpha) = R(\sin x\cos\alpha - \cos x\sin\alpha)\) | ||
| \(\Rightarrow R\cos\alpha = 1,\ R\sin\alpha = \sqrt{3}\) | ||
| \(\Rightarrow R^2 = 1^2 + (\sqrt{3})^2 = 4,\ R = 2\) | B1 | \(R = 2\) |
| \(\tan\alpha = \frac{\sqrt{3}}{1} = \sqrt{3} \Rightarrow \alpha = \frac{\pi}{3}\) | M1 | \(\tan\alpha = \sqrt{3}\) or \(\sin\alpha = \frac{\sqrt{3}}{\text{their }R}\) or \(\cos\alpha = \frac{1}{\text{their }R}\) |
| A1 | \(\alpha = \frac{\pi}{3}\), \(60°\) or \(1.05\) (or better) radians www | |
| \(\Rightarrow \sin x - \sqrt{3}\cos x = 2\sin(x - \frac{\pi}{3})\) | M1 | Using \(x\): their \(\alpha = \frac{\pi}{2}\) or \(90°\), \(\alpha \neq 0\) |
| \(x\) coordinate of P when \(x - \frac{\pi}{3} = \frac{\pi}{2}\) | A1ft | exact radians only (not \(\frac{\pi}{2}\)) |
| \(\Rightarrow x = \frac{5\pi}{6}\) | B1ft | their \(R\) (exact only) |
| \(y = 2\) | ||
| So coordinates are \(\left(\frac{5\pi}{6},\ 2\right)\) | [6] |
## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin x - \sqrt{3}\cos x = R\sin(x-\alpha) = R(\sin x\cos\alpha - \cos x\sin\alpha)$ | | |
| $\Rightarrow R\cos\alpha = 1,\ R\sin\alpha = \sqrt{3}$ | | |
| $\Rightarrow R^2 = 1^2 + (\sqrt{3})^2 = 4,\ R = 2$ | B1 | $R = 2$ |
| $\tan\alpha = \frac{\sqrt{3}}{1} = \sqrt{3} \Rightarrow \alpha = \frac{\pi}{3}$ | M1 | $\tan\alpha = \sqrt{3}$ or $\sin\alpha = \frac{\sqrt{3}}{\text{their }R}$ or $\cos\alpha = \frac{1}{\text{their }R}$ |
| | A1 | $\alpha = \frac{\pi}{3}$, $60°$ or $1.05$ (or better) radians www |
| $\Rightarrow \sin x - \sqrt{3}\cos x = 2\sin(x - \frac{\pi}{3})$ | M1 | Using $x$: their $\alpha = \frac{\pi}{2}$ or $90°$, $\alpha \neq 0$ |
| $x$ coordinate of P when $x - \frac{\pi}{3} = \frac{\pi}{2}$ | A1ft | exact radians only (not $\frac{\pi}{2}$) |
| $\Rightarrow x = \frac{5\pi}{6}$ | B1ft | their $R$ (exact only) |
| $y = 2$ | | |
| So coordinates are $\left(\frac{5\pi}{6},\ 2\right)$ | [6] | |7 Fig. 1 shows part of the graph of $y = \sin x \quad \sqrt { 3 } \cos x$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-3_452_613_1187_745}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Express $\quad \sqrt { } \quad$ in the form $R \sin ( x - \alpha )$, where $R > 0$ and $0 \leqslant \alpha \leqslant \frac { 1 } { 2 } \pi$.\\
Hence write down the exact coordinates of the turning point P .
\hfill \mbox{\textit{OCR C4 Q7 [6]}}