Edexcel Paper 2 Specimen — Question 13 10 marks

Exam BoardEdexcel
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReciprocal Trig & Identities
TypeProve identity then solve equation
DifficultyStandard +0.8 This question requires proving a non-standard reciprocal trig identity using double angle formulas and algebraic manipulation, then applying it to solve an equation involving exact trig values. While the identity proof is moderately challenging, the solution step is straightforward once the identity is established. This is harder than routine A-level questions but accessible to well-prepared students.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals
13. (a) Show that $$\operatorname { cosec } 2 x + \cot 2 x \equiv \cot x , \quad x \neq 90 n ^ { \circ } , n \in \mathbb { Z }$$ (b) Hence, or otherwise, solve, for \(0 \leqslant \theta < 180 ^ { \circ }\), $$\operatorname { cosec } \left( 4 \theta + 10 ^ { \circ } \right) + \cot \left( 4 \theta + 10 ^ { \circ } \right) = \sqrt { 3 }$$ You must show your working.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
Question 13:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\text{cosec}\,2x + \cot 2x = \frac{1}{\sin 2x} + \frac{\cos 2x}{\sin 2x}\)M1 1.2 — Writes both in terms of \(\sin 2x\)
\(= \frac{1+\cos 2x}{\sin 2x}\)M1 1.1b — Combines over common denominator
\(= \frac{1+2\cos^2 x - 1}{2\sin x\cos x} = \frac{2\cos^2 x}{2\sin x\cos x}\)M1, A1 2.1, 1.1b — Applies \(\sin 2x = 2\sin x\cos x\) and double angle formula for numerator
\(= \frac{\cos x}{\sin x} = \cot x\) *A1* 2.1 — Correct proof, no errors in working
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Uses part (a) result with \(x = 2\theta+5°\) to obtain \(\cot(2\theta \pm \ldots°) = \sqrt{3}\)M1 2.2a
\(2\theta + \ldots = 30° \Rightarrow \theta = 12.5°\)M1, A1 1.1b — Applies \(\text{arccot}(\sqrt{3})=30°\) and solves
\(2\theta + 5° = 180° + PV° \Rightarrow \theta = \ldots°\)M1 2.1 — Uses period to find second solution
\(\theta = 102.5°\)A1 1.1b — No extra solutions outside range \(0 \leq \theta < 180°\) 
## Question 13:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{cosec}\,2x + \cot 2x = \frac{1}{\sin 2x} + \frac{\cos 2x}{\sin 2x}$ | M1 | 1.2 — Writes both in terms of $\sin 2x$ |
| $= \frac{1+\cos 2x}{\sin 2x}$ | M1 | 1.1b — Combines over common denominator |
| $= \frac{1+2\cos^2 x - 1}{2\sin x\cos x} = \frac{2\cos^2 x}{2\sin x\cos x}$ | M1, A1 | 2.1, 1.1b — Applies $\sin 2x = 2\sin x\cos x$ and double angle formula for numerator |
| $= \frac{\cos x}{\sin x} = \cot x$ * | A1* | 2.1 — Correct proof, no errors in working |

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses part (a) result with $x = 2\theta+5°$ to obtain $\cot(2\theta \pm \ldots°) = \sqrt{3}$ | M1 | 2.2a |
| $2\theta + \ldots = 30° \Rightarrow \theta = 12.5°$ | M1, A1 | 1.1b — Applies $\text{arccot}(\sqrt{3})=30°$ and solves |
| $2\theta + 5° = 180° + PV° \Rightarrow \theta = \ldots°$ | M1 | 2.1 — Uses period to find second solution |
| $\theta = 102.5°$ | A1 | 1.1b — No extra solutions outside range $0 \leq \theta < 180°$ |
13. (a) Show that

$$\operatorname { cosec } 2 x + \cot 2 x \equiv \cot x , \quad x \neq 90 n ^ { \circ } , n \in \mathbb { Z }$$

(b) Hence, or otherwise, solve, for $0 \leqslant \theta < 180 ^ { \circ }$,

$$\operatorname { cosec } \left( 4 \theta + 10 ^ { \circ } \right) + \cot \left( 4 \theta + 10 ^ { \circ } \right) = \sqrt { 3 }$$

You must show your working.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)

\hfill \mbox{\textit{Edexcel Paper 2  Q13 [10]}}
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