CAIE P3 2023 June — Question 6 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeExpress and solve equation
DifficultyStandard +0.3 This is a standard harmonic form question with straightforward application of compound angle formulas and solving. Part (a) requires expanding cos(x-60°), collecting terms, and using R²=a²+b² with tan α=b/a. Part (b) is a routine substitution (x=2θ) followed by solving R cos(x-α)=2.5. While multi-step, it follows a well-practiced algorithm with no novel insight 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

6
  1. Express \(3 \cos x + 2 \cos \left( x - 60 ^ { \circ } \right)\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). State the exact value of \(R\) and give \(\alpha\) correct to 2 decimal places.
  2. Hence solve the equation $$3 \cos 2 \theta + 2 \cos \left( 2 \theta - 60 ^ { \circ } \right) = 2.5$$ for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

Question 6(b):
AnswerMarks Guidance
AnswerMark Guidance
\(\cos^{-1}\left(\frac{2.5}{R}\right)\)B1 FT SOI [55.0°]. Follow through their \(\sqrt{19}\)
Use correct method to find value of \(2\theta\) (not \(x\)) in interval. Allow sign error in moving \(\alpha\) to right sideM1 \(2\theta = \cos^{-1}\left(\frac{2.5}{R}\right) + 23.41°\) or \(2\theta = 360° - \cos^{-1}\left(\frac{2.5}{R}\right) + 23.41°\) with \(R\) substituted
Obtain one correct answer e.g. \(39.2°\)A1 If working for M1 not seen then M1 implied by \(39.2°\) or \(164.2°\). Must be at least 1 d.p.
Obtain second correct answer e.g. \(164.2°\) and no others in the intervalA1 Must be at least 1 d.p. Ignore answers outside the given interval
## Question 6(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos^{-1}\left(\frac{2.5}{R}\right)$ | **B1 FT** | SOI [55.0°]. Follow through their $\sqrt{19}$ |
| Use correct method to find value of $2\theta$ (not $x$) in interval. Allow sign error in moving $\alpha$ to right side | **M1** | $2\theta = \cos^{-1}\left(\frac{2.5}{R}\right) + 23.41°$ or $2\theta = 360° - \cos^{-1}\left(\frac{2.5}{R}\right) + 23.41°$ with $R$ substituted |
| Obtain one correct answer e.g. $39.2°$ | **A1** | If working for M1 not seen then M1 implied by $39.2°$ or $164.2°$. Must be at least 1 d.p. |
| Obtain second correct answer e.g. $164.2°$ and no others in the interval | **A1** | Must be at least 1 d.p. Ignore answers outside the given interval |

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6
\begin{enumerate}[label=(\alph*)]
\item Express $3 \cos x + 2 \cos \left( x - 60 ^ { \circ } \right)$ in the form $R \cos ( x - \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$. State the exact value of $R$ and give $\alpha$ correct to 2 decimal places.
\item Hence solve the equation

$$3 \cos 2 \theta + 2 \cos \left( 2 \theta - 60 ^ { \circ } \right) = 2.5$$

for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2023 Q6 [8]}}