Question 2 - A-Level Maths

Edexcel C3 — Question 2 10 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Prove identity then solve equation
Difficulty	Standard +0.3 This is a standard C3 trigonometric identity proof followed by a routine equation solve. Part (a) requires straightforward manipulation using double angle formulas and tan = sin/cos. Part (b) uses the proven identity to reduce to a simple equation in cos 2x. The techniques are well-practiced at this level with no novel insight required, making it slightly easier than average.
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

2. (a) Prove that, for $\cos x \neq 0$, $$\sin 2 x - \tan x \equiv \tan x \cos 2 x .$$ (b) Hence, or otherwise, solve the equation $$\sin 2 x - \tan x = 2 \cos 2 x ,$$ for $x$ in the interval $0 \leq x \leq 180 ^ { \circ }$.

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Answer	Marks	Guidance
(a) $\text{LHS} = 2\sin x \cos x - \frac{\sin x}{\cos x}$	M1
$= \frac{2\sin x \cos^2 x - \sin x}{\cos x}$	M1 A1
$= \frac{\sin x(2\cos^2 x - 1)}{\cos x} = \frac{\sin x}{\cos x} \times \cos 2x = \tan x \cos 2x = \text{RHS}$	M1 A1
(b) $\tan x \cos 2x = 2 \cos 2x$	M1
$\cos 2x(\tan x - 2) = 0$	A1
$\cos 2x = 0$ or $\tan x = 2$	B1
$2x = 90, 270$ or $x = 63.4$	M1 A1	(10)
$x = 45°, 63.4°$ (1dp), $135°$

**(a)** $\text{LHS} = 2\sin x \cos x - \frac{\sin x}{\cos x}$ | M1 |
$= \frac{2\sin x \cos^2 x - \sin x}{\cos x}$ | M1 A1 |
$= \frac{\sin x(2\cos^2 x - 1)}{\cos x} = \frac{\sin x}{\cos x} \times \cos 2x = \tan x \cos 2x = \text{RHS}$ | M1 A1 |

**(b)** $\tan x \cos 2x = 2 \cos 2x$ | M1 |
$\cos 2x(\tan x - 2) = 0$ | A1 |
$\cos 2x = 0$ or $\tan x = 2$ | B1 |
$2x = 90, 270$ or $x = 63.4$ | M1 A1 | (10)
$x = 45°, 63.4°$ (1dp), $135°$ |

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2. (a) Prove that, for $\cos x \neq 0$,

$$\sin 2 x - \tan x \equiv \tan x \cos 2 x .$$

(b) Hence, or otherwise, solve the equation

$$\sin 2 x - \tan x = 2 \cos 2 x ,$$

for $x$ in the interval $0 \leq x \leq 180 ^ { \circ }$.\\

\hfill \mbox{\textit{Edexcel C3  Q2 [10]}}

This paper (6 questions)

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Q2 10 Q3 11 Q4 11 Q5 12 Q6 13 Q7 13

Answer	Marks	Guidance
(a) \(\text{LHS} = 2\sin x \cos x - \frac{\sin x}{\cos x}\)	M1
\(= \frac{2\sin x \cos^2 x - \sin x}{\cos x}\)	M1 A1
\(= \frac{\sin x(2\cos^2 x - 1)}{\cos x} = \frac{\sin x}{\cos x} \times \cos 2x = \tan x \cos 2x = \text{RHS}\)	M1 A1
(b) \(\tan x \cos 2x = 2 \cos 2x\)	M1
\(\cos 2x(\tan x - 2) = 0\)	A1
\(\cos 2x = 0\) or \(\tan x = 2\)	B1
\(2x = 90, 270\) or \(x = 63.4\)	M1 A1	(10)
\(x = 45°, 63.4°\) (1dp), \(135°\)