| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Expand then express in harmonic form |
| Difficulty | Standard +0.3 This is a standard harmonic form question with straightforward application of compound angle formulas and solving. Part (i) requires expanding cos(x+45°), collecting terms, and converting to R cos(x+α) form using standard techniques. Part (ii) is routine solving once the harmonic form is established. The question is slightly above average difficulty due to the initial expansion step and numerical precision requirements, but follows a well-practiced procedure with no novel insight required. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use \(\cos(A + B)\) formula to express the given expression in terms of \(\cos x\) and \(\sin x\) | M1 | |
| Collect terms and reach \(\frac{\cos x}{\sqrt{2}} - \frac{3}{\sqrt{2}}\sin x\), or equivalent | A1 | |
| Obtain \(R = 2.236\) | A1 | |
| Use trig formula to find \(\alpha\) | M1 | |
| Obtain \(\alpha = 71.57°\) with no errors seen | A1 | [5] |
| (ii) Evaluate \(\cos^{-1}(2/2.236)\) to at least 1 d.p. (26.56° to 2 d.p., use of \(R = \sqrt{5}\) gives 26.57°) | B1 | |
| Carry out an appropriate method to find a value of \(x\) in the interval \(0° < x < 360°\) | M1 | |
| Obtain answer, e.g. \(x = 315°\) (315.0°) | A1 | |
| Obtain second answer, e.g. 261.9° and no others in the given interval | A1 | [4] |
**(i)** Use $\cos(A + B)$ formula to express the given expression in terms of $\cos x$ and $\sin x$ | M1 |
Collect terms and reach $\frac{\cos x}{\sqrt{2}} - \frac{3}{\sqrt{2}}\sin x$, or equivalent | A1 |
Obtain $R = 2.236$ | A1 |
Use trig formula to find $\alpha$ | M1 |
Obtain $\alpha = 71.57°$ with no errors seen | A1 | [5]
**(ii)** Evaluate $\cos^{-1}(2/2.236)$ to at least 1 d.p. (26.56° to 2 d.p., use of $R = \sqrt{5}$ gives 26.57°) | B1 |
Carry out an appropriate method to find a value of $x$ in the interval $0° < x < 360°$ | M1 |
Obtain answer, e.g. $x = 315°$ (315.0°) | A1 |
Obtain second answer, e.g. 261.9° and no others in the given interval | A1 | [4]
[Ignore answers outside the given range.]
[Treat answers in radians as a misread and deduct A1 from the answers for the angles.]
[SR: Conversion of the equation to a correct quadratic in $\sin x$, $\cos x$, or $\tan x$ earns B1, then M1 for solving a 3-term quadratic and obtaining a value of $x$ in the given interval, and A1 + A1 for the two correct answers (candidates must reject spurious roots to earn the final A1).]\begin{enumerate}[label=(\roman*)]
\item By first expanding $\cos(x + 45°)$, express $\cos(x + 45°) - (\sqrt{2}) \sin x$ in the form $R \cos(x + \alpha)$, where $R > 0$ and $0° < \alpha < 90°$. Give the value of $R$ correct to 4 significant figures and the value of $\alpha$ correct to 2 decimal places. [5]
\item Hence solve the equation
$$\cos(x + 45°) - (\sqrt{2}) \sin x = 2,$$
for $0° < x < 360°$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2013 Q7 [9]}}