CAIE P3 2006 June — Question 4 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2006
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeExpress and solve equation
DifficultyModerate -0.3 This is a standard two-part harmonic form question requiring routine application of the R cos(θ - α) formula (finding R = √(7² + 24²) = 25 and α = arctan(24/7)), followed by solving a straightforward trigonometric equation. While it involves multiple steps and careful arithmetic, it follows a well-practiced textbook procedure with no novel problem-solving required, making it slightly easier than average.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

  1. Express \(7\cos \theta + 24\sin \theta\) in the form \(R\cos(\theta - \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\), giving the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places. [3]
  2. Hence solve the equation $$7\cos \theta + 24\sin \theta = 15,$$ giving all solutions in the interval \(0° \leqslant \theta \leqslant 360°\). [4]

(i)
AnswerMarks Guidance
State answer \(R = 25\)B1
Use trig formula to find \(\alpha\)M1
Obtain \(\alpha = 73.74°\)A1 3
(ii)
AnswerMarks Guidance
Carry out evaluation of \(\cos^{-1}(15/25)\) (\(= 53.1301...\)°)M1
Obtain answer \(126.9°\)A1
Carry out correct method for second answerM1
Obtain answer \(20.6°\) and no others in the range [Ignore answers outside the given range.]A1♦ 4
**(i)**

| State answer $R = 25$ | B1 |
| Use trig formula to find $\alpha$ | M1 |
| Obtain $\alpha = 73.74°$ | A1 | 3 |

**(ii)**

| Carry out evaluation of $\cos^{-1}(15/25)$ ($= 53.1301...$°) | M1 |
| Obtain answer $126.9°$ | A1 |
| Carry out correct method for second answer | M1 |
| Obtain answer $20.6°$ and no others in the range [Ignore answers outside the given range.] | A1♦ | 4 |
\begin{enumerate}[label=(\roman*)]
\item Express $7\cos \theta + 24\sin \theta$ in the form $R\cos(\theta - \alpha)$, where $R > 0$ and $0° < \alpha < 90°$, giving the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places. [3]

\item Hence solve the equation
$$7\cos \theta + 24\sin \theta = 15,$$
giving all solutions in the interval $0° \leqslant \theta \leqslant 360°$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2006 Q4 [7]}}