Question 6 - A-Level Maths

OCR C3 — Question 6 11 marks

Exam Board	OCR
Module	C3 (Core Mathematics 3)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Prove identity then solve equation
Difficulty	Standard +0.8 This question requires proving a non-trivial trigonometric identity involving double angles and reciprocal functions, then solving an equation that combines the identity with cosec². Part (i) demands algebraic manipulation of cot 2x using double angle formulas and common denominators. Part (ii) requires recognizing to substitute the proven identity, applying the Pythagorean identity cot²x + 1 = cosec²x, and solving a resulting quadratic in cot x. While systematic, this goes beyond routine C3 exercises and requires strong algebraic fluency with multiple trig identities.
Spec	1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 1.05l Double angle formulae: and compound angle formulae 1.05o Trigonometric equations: solve in given intervals 1.05p Proof involving trig: functions and identities

6. (i) Prove the identity $$2 \cot 2 x + \tan x \equiv \cot x , \quad x \neq \frac { n } { 2 } \pi , \quad n \in \mathbb { Z }$$ (ii) Solve, for $0 \leq x < \pi$, the equation $$2 \cot 2 x + \tan x = \operatorname { cosec } ^ { 2 } x - 7$$ giving your answers to 2 decimal places.

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Question 6:

Part (i):

Answer	Marks	Guidance
Working/Answer	Mark	Notes
$\text{LHS} \equiv \frac{2\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x}$	M1
$\equiv \frac{\cos 2x}{\sin x \cos x} + \frac{\sin x}{\cos x}$	M1
$\equiv \frac{\cos 2x + \sin^2 x}{\sin x \cos x}$	A1
$\equiv \frac{(\cos^2 x - \sin^2 x) + \sin^2 x}{\sin x \cos x}$	M1
$\equiv \frac{\cos^2 x}{\sin x \cos x} \equiv \frac{\cos x}{\sin x} \equiv \cot x \equiv \text{RHS}$	A1

Part (ii):

Answer	Marks	Guidance
Working/Answer	Mark	Notes
$\cot x = \text{cosec}^2 x - 7$, using $\cot x = 1 + \cot^2 x - 7$	M1
$\cot^2 x - \cot x - 6 = 0$, $(\cot x + 2)(\cot x - 3) = 0$	M1
$\cot x = -2$ or $3$	A1
$\tan x = -\frac{1}{2}$ or $\frac{1}{3}$	M1
$x = \pi - 0.4636$ or $0.32$
$x = 0.32, \ 2.68$ (2dp)	A2	(11)

# Question 6:

## Part (i):

| Working/Answer | Mark | Notes |
|---|---|---|
| $\text{LHS} \equiv \frac{2\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x}$ | M1 | |
| $\equiv \frac{\cos 2x}{\sin x \cos x} + \frac{\sin x}{\cos x}$ | M1 | |
| $\equiv \frac{\cos 2x + \sin^2 x}{\sin x \cos x}$ | A1 | |
| $\equiv \frac{(\cos^2 x - \sin^2 x) + \sin^2 x}{\sin x \cos x}$ | M1 | |
| $\equiv \frac{\cos^2 x}{\sin x \cos x} \equiv \frac{\cos x}{\sin x} \equiv \cot x \equiv \text{RHS}$ | A1 | |

## Part (ii):

| Working/Answer | Mark | Notes |
|---|---|---|
| $\cot x = \text{cosec}^2 x - 7$, using $\cot x = 1 + \cot^2 x - 7$ | M1 | |
| $\cot^2 x - \cot x - 6 = 0$, $(\cot x + 2)(\cot x - 3) = 0$ | M1 | |
| $\cot x = -2$ or $3$ | A1 | |
| $\tan x = -\frac{1}{2}$ or $\frac{1}{3}$ | M1 | |
| $x = \pi - 0.4636$ or $0.32$ | | |
| $x = 0.32, \ 2.68$ (2dp) | A2 | **(11)** |

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6. (i) Prove the identity

$$2 \cot 2 x + \tan x \equiv \cot x , \quad x \neq \frac { n } { 2 } \pi , \quad n \in \mathbb { Z }$$

(ii) Solve, for $0 \leq x < \pi$, the equation

$$2 \cot 2 x + \tan x = \operatorname { cosec } ^ { 2 } x - 7$$

giving your answers to 2 decimal places.\\

\hfill \mbox{\textit{OCR C3  Q6 [11]}}

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Q1 6 Q2 7 Q3 8 Q4 8 Q5 8 Q6 11 Q7 11 Q8 13

Answer	Marks	Guidance
Working/Answer	Mark	Notes
\(\cot x = \text{cosec}^2 x - 7\), using \(\cot x = 1 + \cot^2 x - 7\)	M1
\(\cot^2 x - \cot x - 6 = 0\), \((\cot x + 2)(\cot x - 3) = 0\)	M1
\(\cot x = -2\) or \(3\)	A1
\(\tan x = -\frac{1}{2}\) or \(\frac{1}{3}\)	M1
\(x = \pi - 0.4636\) or \(0.32\)
\(x = 0.32, \ 2.68\) (2dp)	A2	(11)