Question 7 - A-Level Maths

OCR C3 — Question 7 9 marks

Exam Board	OCR
Module	C3 (Core Mathematics 3)
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Addition & Double Angle Formulae
Type	Prove identity then solve equation
Difficulty	Standard +0.3 Part (i) is a straightforward identity proof using standard double angle formulae (sin 2x = 2sin x cos x, cos 2x) and tan x = sin x/cos x. Part (ii) applies the proven identity to solve a trigonometric equation, requiring substitution and solving a quadratic in cos 2x, followed by finding angles in a given range. This is slightly above average due to the two-part structure and the need to manipulate the equation after substitution, but remains a standard C3 exercise with well-practiced techniques.
Spec	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

7. (i) Prove that, for $\cos x \neq 0$, $$\sin 2 x - \tan x \equiv \tan x \cos 2 x$$ (ii) 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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Question 7:

Part (i)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
LHS $\equiv 2\sin x \cos x - \frac{\sin x}{\cos x}$	M1
$\equiv \frac{2\sin x \cos^2 x - \sin x}{\cos x}$	M1 A1
$\equiv \frac{\sin x(2\cos^2 x - 1)}{\cos x} \equiv \frac{\sin x}{\cos x} \times \cos 2x \equiv \tan x \cos 2x \equiv$ RHS	M1 A1

Part (ii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\tan x \cos 2x = 2\cos 2x$
$\cos 2x(\tan x - 2) = 0$	M1
$\cos 2x = 0$ or $\tan x = 2$	A1
$2x = 90, 270$ or $x = 63.4$
$x = 45°, 63.4°$ (3sf), $135°$	A2	(9)

# Question 7:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| LHS $\equiv 2\sin x \cos x - \frac{\sin x}{\cos x}$ | M1 | |
| $\equiv \frac{2\sin x \cos^2 x - \sin x}{\cos x}$ | M1 A1 | |
| $\equiv \frac{\sin x(2\cos^2 x - 1)}{\cos x} \equiv \frac{\sin x}{\cos x} \times \cos 2x \equiv \tan x \cos 2x \equiv$ RHS | M1 A1 | |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan x \cos 2x = 2\cos 2x$ | | |
| $\cos 2x(\tan x - 2) = 0$ | M1 | |
| $\cos 2x = 0$ or $\tan x = 2$ | A1 | |
| $2x = 90, 270$ or $x = 63.4$ | | |
| $x = 45°, 63.4°$ (3sf), $135°$ | A2 | **(9)** |

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7. (i) Prove that, for $\cos x \neq 0$,

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

(ii) 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{OCR C3  Q7 [9]}}

This paper (9 questions)

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Q1 5 Q2 6 Q3 6 Q4 7 Q5 8 Q6 9 Q7 9 Q8 11 Q9 11

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\tan x \cos 2x = 2\cos 2x\)
\(\cos 2x(\tan x - 2) = 0\)	M1
\(\cos 2x = 0\) or \(\tan x = 2\)	A1
\(2x = 90, 270\) or \(x = 63.4\)
\(x = 45°, 63.4°\) (3sf), \(135°\)	A2	(9)