Question 6 - A-Level Maths

Edexcel C3 — Question 6 11 marks

Exam Board	Edexcel
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 Part (a) requires proving a non-trivial trigonometric identity involving cot 2x and double angle formulas, demanding algebraic manipulation beyond standard identities. Part (b) combines the proven identity with cosec²x, requiring substitution, use of Pythagorean identities, and solving a resulting equation numerically within a restricted domain—this is more demanding than routine C3 trigonometric equations.
Spec	1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

6. (a) Prove the identity $$2 \cot 2 x + \tan x \equiv \cot x , \quad x \neq \frac { n } { 2 } \pi , \quad n \in \mathbb { Z } .$$ (b) 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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(a)

Answer	Marks
$\text{LHS} = \frac{2\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x}$	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
$\equiv \frac{\cos^2 x}{\sin x \cos x}$
$\equiv \frac{\cos x}{\sin x}$
$\equiv \cot x = \text{RHS}$	A1

(b)

Answer	Marks	Guidance
$\cot x = \cosec^2 x - 7$	M1
$\cot x = 1 + \cot^2 x - 7$	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
$x = 0.32, 2.68$ (2dp)	A2	(11 marks)

**(a)**
$\text{LHS} = \frac{2\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x}$ | 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 |
$\equiv \frac{\cos^2 x}{\sin x \cos x}$ | |
$\equiv \frac{\cos x}{\sin x}$ | |
$\equiv \cot x = \text{RHS}$ | A1 |

**(b)**
$\cot x = \cosec^2 x - 7$ | M1 |
$\cot x = 1 + \cot^2 x - 7$ | 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 |
$x = 0.32, 2.68$ (2dp) | A2 | (11 marks)

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6. (a) Prove the identity

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

(b) 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{Edexcel C3  Q6 [11]}}

This paper (7 questions)

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Q1 8 Q2 9 Q3 10 Q4 11 Q5 11 Q6 11 Q7 15

Answer	Marks
\(\text{LHS} = \frac{2\cos 2x}{\sin 2x} + \frac{\sin x}{\cos x}\)	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
\(\equiv \frac{\cos^2 x}{\sin x \cos x}\)
\(\equiv \frac{\cos x}{\sin x}\)
\(\equiv \cot x = \text{RHS}\)	A1

Answer	Marks	Guidance
\(\cot x = \cosec^2 x - 7\)	M1
\(\cot x = 1 + \cot^2 x - 7\)	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
\(x = 0.32, 2.68\) (2dp)	A2	(11 marks)