CAIE P2 2010 November — Question 6 7 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2010
SessionNovember
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 sin(θ - α) formula (finding R = √5 and α ≈ 26.57°) followed by solving a straightforward equation using inverse sine. While it involves multiple steps, the techniques are well-practiced and follow a predictable template 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

6
  1. Express \(2 \sin \theta - \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
  2. Hence solve the equation $$2 \sin \theta - \cos \theta = - 0.4$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

(i)
AnswerMarks Guidance
State \(R = \sqrt{5}\)B1
Use trig formula to find \(\alpha\)M1
Obtain \(\alpha = 26.57°\) with no errors seenA1 [3]
(ii)
AnswerMarks Guidance
Carry out evaluation of \(\sin\left(\frac{\pm 0.4}{\sqrt{5}}\right)\) (≈ ±10.3048°)M1
Obtain answer 16.3°A1
Carry out correct method for second answerM1
Obtain answer 216.9° and no others in the rangeA1 [4]
**(i)**

| State $R = \sqrt{5}$ | B1 |
| Use trig formula to find $\alpha$ | M1 |
| Obtain $\alpha = 26.57°$ with no errors seen | A1 | [3] |

**(ii)**

| Carry out evaluation of $\sin\left(\frac{\pm 0.4}{\sqrt{5}}\right)$ (≈ ±10.3048°) | M1 |
| Obtain answer 16.3° | A1 |
| Carry out correct method for second answer | M1 |
| Obtain answer 216.9° and no others in the range | A1 | [4] |
6 (i) Express $2 \sin \theta - \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$, giving the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.\\
(ii) Hence solve the equation

$$2 \sin \theta - \cos \theta = - 0.4$$

giving all solutions in the interval $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P2 2010 Q6 [7]}}