| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 Part (a) is a standard harmonic form conversion using R cos(θ-α) = R cos α cos θ + R sin α sin θ, requiring routine application of Pythagorean identity and inverse tangent. Part (b) applies this to solve a straightforward equation with a double angle substitution. This is a textbook exercise testing standard technique with minimal problem-solving demand, making it slightly easier than average. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State \(R = 13\) | B1 | Allow if \(\sqrt{12^2 + (-5)^2}\) seen |
| Use correct trig formulae to find \(\alpha = \tan^{-1}(\pm 5/12) = \cos^{-1}(\pm 12/13) = \sin^{-1}(\pm 5/13)\) | M1 | \(\cos(\alpha) = 12\) and \(\sin(\alpha) = 5\) M0; however \(\sin(\alpha)/\cos(\alpha) = 5/12\) or \(-5/12\) with no error seen, or \(\tan(\alpha) = 5/12\) or \(-5/12\) quoted then allow |
| Obtain \(\alpha = 0.395\) | A1 | CWO if negative sign seen when finding \(R\) then A0; if degrees 22.6 A0 MR; only penalise degrees once in (a) and (b); note \(\alpha = 0.39479\ldots\) |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos^{-1}\left(\frac{6}{R}\right)\) | B1FT | SOI; \(1.0910\ldots\) FT *their* incorrect \(R\) |
| Use correct method to find a value of \(2x\) in the interval | M1 | \(2x = \cos^{-1}\left(\frac{6}{R}\right) + \alpha\) or \(2\pi - \cos^{-1}\left(\frac{6}{R}\right) + \alpha\); allow if \(\cos(2x + 0.395)\) seen |
| Obtain answer e.g. \(x = 0.743\) or \(0.742\) | A1 | 42.5 or 42.6 degrees |
| Obtain second answer e.g. \(x = 2.79\) and no others in the interval | A1 | 159.8, 159.9 or 160.0 degrees all possible depending on 3dp or 4dp |
| Total | 4 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State $R = 13$ | **B1** | Allow if $\sqrt{12^2 + (-5)^2}$ seen |
| Use correct trig formulae to find $\alpha = \tan^{-1}(\pm 5/12) = \cos^{-1}(\pm 12/13) = \sin^{-1}(\pm 5/13)$ | **M1** | $\cos(\alpha) = 12$ and $\sin(\alpha) = 5$ M0; however $\sin(\alpha)/\cos(\alpha) = 5/12$ or $-5/12$ with no error seen, or $\tan(\alpha) = 5/12$ or $-5/12$ quoted then allow |
| Obtain $\alpha = 0.395$ | **A1** | CWO if negative sign seen when finding $R$ then A0; if degrees 22.6 A0 MR; only penalise degrees once in **(a)** and **(b)**; note $\alpha = 0.39479\ldots$ |
| **Total** | **3** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos^{-1}\left(\frac{6}{R}\right)$ | **B1FT** | SOI; $1.0910\ldots$ FT *their* incorrect $R$ |
| Use correct method to find a value of $2x$ in the interval | **M1** | $2x = \cos^{-1}\left(\frac{6}{R}\right) + \alpha$ or $2\pi - \cos^{-1}\left(\frac{6}{R}\right) + \alpha$; allow if $\cos(2x + 0.395)$ seen |
| Obtain answer e.g. $x = 0.743$ or $0.742$ | **A1** | 42.5 or 42.6 degrees |
| Obtain second answer e.g. $x = 2.79$ and no others in the interval | **A1** | 159.8, 159.9 or 160.0 degrees all possible depending on 3dp or 4dp |
| **Total** | **4** | |6
\begin{enumerate}[label=(\alph*)]
\item Express $5 \sin \theta + 12 \cos \theta$ in the form $R \cos ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$.
\item Hence solve the equation $5 \sin 2 x + 12 \cos 2 x = 6$ for $0 \leqslant x \leqslant \pi$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q6 [7]}}