Question 7 - A-Level Maths

Edexcel P2 2020 January — Question 7 7 marks

Exam Board	Edexcel
Module	P2 (Pure Mathematics 2)
Year	2020
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Reciprocal Trig & Identities
Type	Convert equation to quadratic form
Difficulty	Standard +0.3 This is a standard A-level technique question requiring conversion of a trigonometric equation to quadratic form using tan θ = sin θ/cos θ and the Pythagorean identity. Part (a) is routine algebraic manipulation (showing a given result), and part (b) applies the same method with a double angle substitution. The steps are well-practiced and require no novel insight, making it slightly easier than average.
Spec	1.02f Solve quadratic equations: including in a function of unknown 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

7. (a) Show that the equation $$8 \tan \theta = 3 \cos \theta$$ may be rewritten in the form $$3 \sin ^ { 2 } \theta + 8 \sin \theta - 3 = 0$$ (b) Hence solve, for $0 \leqslant x \leqslant 90 ^ { \circ }$, the equation $$8 \tan 2 x = 3 \cos 2 x$$ giving your answers to 2 decimal places.

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Answer	Marks
(a) $8\tan\theta = 3\cos\theta$	—
Uses $\tan\theta = \frac{\sin\theta}{\cos\theta} \Rightarrow 8\frac{\sin\theta}{\cos\theta} = 3\cos\theta$	M1
$8\sin\theta = 3\cos^2\theta$ oe	M1
Uses $\cos^2\theta = 1 - \sin^2\theta \Rightarrow 8\sin\theta = 3(1 - \sin^2\theta)$	—
$3\sin^2\theta + 8\sin\theta - 3 = 0$ *	A1

(3 marks)

Answer	Marks
(b) $(3\sin^2x - 1)(\sin^2x + 3) = 0$	M1
Critical value(s) of $\sin^2x = \frac{1}{3}$, $(\sin^2x = -3)$	A1

(a) $8\tan\theta = 3\cos\theta$ | —

Uses $\tan\theta = \frac{\sin\theta}{\cos\theta} \Rightarrow 8\frac{\sin\theta}{\cos\theta} = 3\cos\theta$ | M1

$8\sin\theta = 3\cos^2\theta$ oe | M1

Uses $\cos^2\theta = 1 - \sin^2\theta \Rightarrow 8\sin\theta = 3(1 - \sin^2\theta)$ | —

$3\sin^2\theta + 8\sin\theta - 3 = 0$ * | A1

(3 marks)

(b) $(3\sin^2x - 1)(\sin^2x + 3) = 0$ | M1

Critical value(s) of $\sin^2x = \frac{1}{3}$, $(\sin^2x = -3)$ | A1

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7. (a) Show that the equation

$$8 \tan \theta = 3 \cos \theta$$

may be rewritten in the form

$$3 \sin ^ { 2 } \theta + 8 \sin \theta - 3 = 0$$

(b) Hence solve, for $0 \leqslant x \leqslant 90 ^ { \circ }$, the equation

$$8 \tan 2 x = 3 \cos 2 x$$

giving your answers to 2 decimal places.\\

\hfill \mbox{\textit{Edexcel P2 2020 Q7 [7]}}

This paper (10 questions)

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Q1 7 Q2 7 Q3 8 Q4 6 Q5 8 Q6 8 Q7 7 Q8 7 Q9 7 Q10 10

Answer	Marks
(a) \(8\tan\theta = 3\cos\theta\)	—
Uses \(\tan\theta = \frac{\sin\theta}{\cos\theta} \Rightarrow 8\frac{\sin\theta}{\cos\theta} = 3\cos\theta\)	M1
\(8\sin\theta = 3\cos^2\theta\) oe	M1
Uses \(\cos^2\theta = 1 - \sin^2\theta \Rightarrow 8\sin\theta = 3(1 - \sin^2\theta)\)	—
\(3\sin^2\theta + 8\sin\theta - 3 = 0\) *	A1

Answer	Marks
(b) \((3\sin^2x - 1)(\sin^2x + 3) = 0\)	M1
Critical value(s) of \(\sin^2x = \frac{1}{3}\), \((\sin^2x = -3)\)	A1