Question 5 - A-Level Maths

Edexcel C3 — Question 5 12 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Harmonic Form
Type	Express and solve equation
Difficulty	Moderate -0.3 This is a standard C3 harmonic form question following a predictable three-part structure: (a) convert to R sin(θ+α) using standard formulas, (b) state maximum from R and solve simple equation, (c) solve a straightforward equation using the result from (a). All parts use routine techniques with no novel problem-solving required, making it slightly easier than average.
Spec	1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc 1.05o Trigonometric equations: solve in given intervals

5. (a) Express $\sqrt { 3 } \sin \theta + \cos \theta$ in the form $R \sin ( \theta + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.
(b) State the maximum value of $\sqrt { 3 } \sin \theta + \cos \theta$ and the smallest positive value of $\theta$ for which this maximum value occurs.
(c) Solve the equation $$\sqrt { 3 } \sin \theta + \cos \theta + \sqrt { 3 } = 0 ,$$ for $\theta$ in the interval $- \pi \leq \theta \leq \pi$, giving your answers in terms of $\pi$.

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Answer	Marks	Guidance
(a) $\sqrt{3}\sin\theta + \cos\theta = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$	M1 A1
$R\cos\alpha = \sqrt{3}, R\sin\alpha = 1$ $\therefore R = \sqrt{3+1} = 2$	M1 A1
$\tan\alpha = \frac{1}{\sqrt{3}}, \alpha = \frac{\pi}{6}$ $\therefore \sqrt{3}\sin\theta + \cos\theta = 2\sin(\theta + \frac{\pi}{6})$	M1 A1
(b) maximum $= 2$	B1
occurs when $\theta + \frac{\pi}{6} = \frac{\pi}{2}, \theta = \frac{\pi}{3}$	M1 A1
(c) $2\sin(\theta + \frac{\pi}{6}) + \sqrt{3} = 0, \sin(\theta + \frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$	M1
$\theta + \frac{\pi}{6} = -\frac{\pi}{3}, -\pi + \frac{\pi}{3} = -\frac{\pi}{3}, -\frac{2\pi}{3}$	B1 M1
$\theta = -\frac{5\pi}{6}, -\frac{\pi}{2}$	A2	(12)

**(a)** $\sqrt{3}\sin\theta + \cos\theta = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$ | M1 A1 |
$R\cos\alpha = \sqrt{3}, R\sin\alpha = 1$ $\therefore R = \sqrt{3+1} = 2$ | M1 A1 |
$\tan\alpha = \frac{1}{\sqrt{3}}, \alpha = \frac{\pi}{6}$ $\therefore \sqrt{3}\sin\theta + \cos\theta = 2\sin(\theta + \frac{\pi}{6})$ | M1 A1 |

**(b)** maximum $= 2$ | B1 |
occurs when $\theta + \frac{\pi}{6} = \frac{\pi}{2}, \theta = \frac{\pi}{3}$ | M1 A1 |

**(c)** $2\sin(\theta + \frac{\pi}{6}) + \sqrt{3} = 0, \sin(\theta + \frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$ | M1 |
$\theta + \frac{\pi}{6} = -\frac{\pi}{3}, -\pi + \frac{\pi}{3} = -\frac{\pi}{3}, -\frac{2\pi}{3}$ | B1 M1 |
$\theta = -\frac{5\pi}{6}, -\frac{\pi}{2}$ | A2 | (12)

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5. (a) Express $\sqrt { 3 } \sin \theta + \cos \theta$ in the form $R \sin ( \theta + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.\\
(b) State the maximum value of $\sqrt { 3 } \sin \theta + \cos \theta$ and the smallest positive value of $\theta$ for which this maximum value occurs.\\
(c) Solve the equation

$$\sqrt { 3 } \sin \theta + \cos \theta + \sqrt { 3 } = 0 ,$$

for $\theta$ in the interval $- \pi \leq \theta \leq \pi$, giving your answers in terms of $\pi$.\\

\hfill \mbox{\textit{Edexcel C3  Q5 [12]}}

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

Answer	Marks	Guidance
(a) \(\sqrt{3}\sin\theta + \cos\theta = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha\)	M1 A1
\(R\cos\alpha = \sqrt{3}, R\sin\alpha = 1\) \(\therefore R = \sqrt{3+1} = 2\)	M1 A1
\(\tan\alpha = \frac{1}{\sqrt{3}}, \alpha = \frac{\pi}{6}\) \(\therefore \sqrt{3}\sin\theta + \cos\theta = 2\sin(\theta + \frac{\pi}{6})\)	M1 A1
(b) maximum \(= 2\)	B1
occurs when \(\theta + \frac{\pi}{6} = \frac{\pi}{2}, \theta = \frac{\pi}{3}\)	M1 A1
(c) \(2\sin(\theta + \frac{\pi}{6}) + \sqrt{3} = 0, \sin(\theta + \frac{\pi}{6}) = -\frac{\sqrt{3}}{2}\)	M1
\(\theta + \frac{\pi}{6} = -\frac{\pi}{3}, -\pi + \frac{\pi}{3} = -\frac{\pi}{3}, -\frac{2\pi}{3}\)	B1 M1
\(\theta = -\frac{5\pi}{6}, -\frac{\pi}{2}\)	A2	(12)