CAIE P3 2002 November — Question 5 8 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2002
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeExpress and solve equation
DifficultyStandard +0.3 This is a standard harmonic form question requiring routine application of R sin(θ - α) transformation, solving a basic trigonometric equation, and finding a maximum value. All three parts follow textbook procedures with no novel insight required, making it slightly easier than average for A-level.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

5
  1. Express \(4 \sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), stating the value of \(\alpha\) correct to 2 decimal places. Hence
  2. solve the equation $$4 \sin \theta - 3 \cos \theta = 2$$ giving all values of \(\theta\) such that \(0 ^ { \circ } < \theta < 360 ^ { \circ }\),
  3. write down the greatest value of \(\frac { 1 } { 4 \sin \theta - 3 \cos \theta + 6 }\).

AnswerMarks Guidance
ContentMark Guidance
(i) State or imply at any stage that \(R = 5\)B1
Use trig formula to find \(\alpha\)M1
Obtain answer \(\alpha = 36.87°\)A1 Max 3 marks
(ii) EITHER: Carry out, or indicate need for, calculation of \(\sin^{-1}(\frac{3}{5})\)M1
Obtain answer \(60.4°\) (or \(60.5°\))A1
Carry out correct method for second root i.e. \(180° - 23.578° + 36.870°\)M1
Obtain answer \(193.3°\) and no others in rangeA1
OR: Obtain a three-term quadratic equation in \(\sin\theta\) or \(\cos\theta\)M1
Solve a two- or three-term quadratic and calculate an angleM1
Obtain answer \(60.4°\) (or \(60.5°\))A1
Obtain answer \(193.3°\) and no others in rangeA1 Max 4 marks
(iii) State greatest value is \(1\)B1✓ Treat work in radians as a misread, scoring a maximum of 7. The angles are 0.644, 1.06 and 3.37.
| Content | Mark | Guidance |
|---------|------|----------|
| **(i)** State or imply at any stage that $R = 5$ | B1 | |
| Use trig formula to find $\alpha$ | M1 | |
| Obtain answer $\alpha = 36.87°$ | A1 | Max 3 marks |
| **(ii)** **EITHER:** Carry out, or indicate need for, calculation of $\sin^{-1}(\frac{3}{5})$ | M1 | |
| Obtain answer $60.4°$ (or $60.5°$) | A1 | |
| Carry out correct method for second root i.e. $180° - 23.578° + 36.870°$ | M1 | |
| Obtain answer $193.3°$ and no others in range | A1 | |
| **OR:** Obtain a three-term quadratic equation in $\sin\theta$ or $\cos\theta$ | M1 | |
| Solve a two- or three-term quadratic and calculate an angle | M1 | |
| Obtain answer $60.4°$ (or $60.5°$) | A1 | |
| Obtain answer $193.3°$ and no others in range | A1 | Max 4 marks |
| **(iii)** State greatest value is $1$ | B1✓ | Treat work in radians as a misread, scoring a maximum of 7. The angles are 0.644, 1.06 and 3.37. |Max 1 mark |

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5 (i) Express $4 \sin \theta - 3 \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$, stating the value of $\alpha$ correct to 2 decimal places.

Hence\\
(ii) solve the equation

$$4 \sin \theta - 3 \cos \theta = 2$$

giving all values of $\theta$ such that $0 ^ { \circ } < \theta < 360 ^ { \circ }$,\\
(iii) write down the greatest value of $\frac { 1 } { 4 \sin \theta - 3 \cos \theta + 6 }$.

\hfill \mbox{\textit{CAIE P3 2002 Q5 [8]}}