CAIE P2 2020 November — Question 1 3 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2020
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeSolve equation with reciprocal functions
DifficultyModerate -0.8 This is a straightforward reciprocal trig equation requiring only basic manipulation: rewrite as 7cos θ/sin θ = 3/sin θ, cancel sin θ, then solve 7cos θ = 3. Single-step algebraic manipulation with standard inverse trig to finish, simpler than typical A-level questions.
Spec1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05o Trigonometric equations: solve in given intervals

1 Solve the equation \(7 \cot \theta = 3 \operatorname { cosec } \theta\) for \(0 ^ { \circ } < \theta < 90 ^ { \circ }\).

Question 1:
AnswerMarks Guidance
AnswerMark Guidance
Use \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\) and \(\cosec\theta = \dfrac{1}{\sin\theta}\)B1 SOI
Simplify to obtain \(\cos\theta = k\) where \(0 < k < 1\)M1
Obtain \(\cos\theta = \dfrac{3}{7}\) and hence \(\theta = 64.6\) and no other solutions in the rangeA1
Alternative method for question 1:
AnswerMarks Guidance
AnswerMark Guidance
Use identity \(\cosec^2\theta = 1 + \cot^2\theta\)B1
Simplify to obtain \(\tan\theta = k_1\) or \(\sin\theta = k_2\) where \(0 < k_2 < 1\)M1
Obtain \(\tan\theta = \dfrac{1}{3}\sqrt{40}\) or \(\sin\theta = \dfrac{1}{7}\sqrt{40}\) and hence \(\theta = 64.6\) and no other solutions in the rangeA1
Total3
## Question 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| Use $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$ and $\cosec\theta = \dfrac{1}{\sin\theta}$ | B1 | SOI |
| Simplify to obtain $\cos\theta = k$ where $0 < k < 1$ | M1 | |
| Obtain $\cos\theta = \dfrac{3}{7}$ and hence $\theta = 64.6$ and no other solutions in the range | A1 | |

**Alternative method for question 1:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use identity $\cosec^2\theta = 1 + \cot^2\theta$ | B1 | |
| Simplify to obtain $\tan\theta = k_1$ or $\sin\theta = k_2$ where $0 < k_2 < 1$ | M1 | |
| Obtain $\tan\theta = \dfrac{1}{3}\sqrt{40}$ or $\sin\theta = \dfrac{1}{7}\sqrt{40}$ and hence $\theta = 64.6$ and no other solutions in the range | A1 | |
| **Total** | **3** | |
1 Solve the equation $7 \cot \theta = 3 \operatorname { cosec } \theta$ for $0 ^ { \circ } < \theta < 90 ^ { \circ }$.\\

\hfill \mbox{\textit{CAIE P2 2020 Q1 [3]}}