Question 1 - A-Level Maths

Edexcel C3 2007 January — Question 1 7 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2007
Session	January
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Addition & Double Angle Formulae
Type	Given sin/cos/tan, find other expressions
Difficulty	Moderate -0.3 Part (a) is a standard bookwork derivation using addition and double angle formulae with clear guidance on the method. Part (b) is straightforward substitution into the derived formula. This is a routine C3 question testing formula manipulation rather than problem-solving, making it slightly easier than average.
Spec	1.01a Proof: structure of mathematical proof and logical steps 1.05l Double angle formulae: and compound angle formulae

(a) By writing $\sin 3 \theta$ as $\sin ( 2 \theta + \theta )$, show that

$$\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$$ (b) Given that $\sin \theta = \frac { \sqrt { } 3 } { 4 }$, find the exact value of $\sin 3 \theta$.

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Answer	Marks	Guidance
(a) $\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta = 2\sin\theta\cos^2\theta + (1-2\sin^2\theta)\sin\theta = 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta$	B1, B1, M1, A1	cso (5 marks)
(b) $\sin 3\theta = 3x\frac{\sqrt{3}}{4} - 4\left(\frac{\sqrt{3}}{4}\right)^3 = \frac{3\sqrt{3}}{4} - \frac{3\sqrt{3}}{16} = \frac{9\sqrt{3}}{16}$ or exact equivalent	M1, A1	(2 marks)

(a) $\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta = 2\sin\theta\cos^2\theta + (1-2\sin^2\theta)\sin\theta = 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta$ | B1, B1, M1, A1 | cso (5 marks)

(b) $\sin 3\theta = 3x\frac{\sqrt{3}}{4} - 4\left(\frac{\sqrt{3}}{4}\right)^3 = \frac{3\sqrt{3}}{4} - \frac{3\sqrt{3}}{16} = \frac{9\sqrt{3}}{16}$ or exact equivalent | M1, A1 | (2 marks)

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\begin{enumerate}
  \item (a) By writing $\sin 3 \theta$ as $\sin ( 2 \theta + \theta )$, show that
\end{enumerate}

$$\sin 3 \theta = 3 \sin \theta - 4 \sin ^ { 3 } \theta$$

(b) Given that $\sin \theta = \frac { \sqrt { } 3 } { 4 }$, find the exact value of $\sin 3 \theta$.\\

\hfill \mbox{\textit{Edexcel C3 2007 Q1 [7]}}

This paper (8 questions)

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Q1 7 Q2 8 Q3 9 Q4 11 Q5 8 Q6 13 Q7 13 Q8 6

Answer	Marks	Guidance
(a) \(\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta = 2\sin\theta\cos^2\theta + (1-2\sin^2\theta)\sin\theta = 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta\)	B1, B1, M1, A1	cso (5 marks)
(b) \(\sin 3\theta = 3x\frac{\sqrt{3}}{4} - 4\left(\frac{\sqrt{3}}{4}\right)^3 = \frac{3\sqrt{3}}{4} - \frac{3\sqrt{3}}{16} = \frac{9\sqrt{3}}{16}\) or exact equivalent	M1, A1	(2 marks)