Question 7 - A-Level Maths

Edexcel C3 — Question 7 13 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Harmonic Form
Type	Express and solve equation
Difficulty	Standard +0.3 This is a standard C3 harmonic form question with routine steps: finding R and α using Pythagorean theorem and arctan, stating minimum value (-R), and solving an equation using the harmonic form. All techniques are textbook exercises with no novel insight 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

7. (a) Express $4 \sin x + 3 \cos x$ in the form $R \sin ( x + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.
(b) State the minimum value of $4 \sin x + 3 \cos x$ and the smallest positive value of $x$ for which this minimum value occurs.
(c) Solve the equation $$4 \sin 2 \theta + 3 \cos 2 \theta = 2$$ for $\theta$ in the interval $0 \leq \theta \leq \pi$, giving your answers to 2 decimal places.

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Answer	Marks	Guidance
(a) $4 \sin x + 3 \cos x = R \sin x \cos \alpha + R \cos x \sin \alpha$	M1
$R \cos \alpha = 4$, $R \sin \alpha = 3$	M1
$\therefore R = \sqrt{4^2 + 3^2} = 5$	M1 A1
$\tan \alpha = \frac{3}{4}$, $\alpha = 0.644$ (3sf)	M1 A1
$\therefore 4 \sin x + 3 \cos x = 5 \sin (x + 0.644)$	M1
(b) minimum $= -5$	B1
occurs when $x + 0.6435 = \frac{3\pi}{2}$, $x = 4.07$ (3sf)	M1 A1
(c) $5 \sin (2\theta + 0.6435) = 2$	M1
$\sin (2\theta + 0.6435) = 0.4$	M1
$2\theta + 0.6435 = \pi - 0.4115, 2\pi + 0.4115$	B1 M1
$2\theta = 2.087, 6.051$	M1
$\theta = 1.04, 3.03$ (2dp)	A2	(13 marks)

**(a)** $4 \sin x + 3 \cos x = R \sin x \cos \alpha + R \cos x \sin \alpha$ | M1 |

$R \cos \alpha = 4$, $R \sin \alpha = 3$ | M1 |

$\therefore R = \sqrt{4^2 + 3^2} = 5$ | M1 A1 |

$\tan \alpha = \frac{3}{4}$, $\alpha = 0.644$ (3sf) | M1 A1 |

$\therefore 4 \sin x + 3 \cos x = 5 \sin (x + 0.644)$ | M1 |

**(b)** minimum $= -5$ | B1 |

occurs when $x + 0.6435 = \frac{3\pi}{2}$, $x = 4.07$ (3sf) | M1 A1 |

**(c)** $5 \sin (2\theta + 0.6435) = 2$ | M1 |

$\sin (2\theta + 0.6435) = 0.4$ | M1 |

$2\theta + 0.6435 = \pi - 0.4115, 2\pi + 0.4115$ | B1 M1 |

$2\theta = 2.087, 6.051$ | M1 |

$\theta = 1.04, 3.03$ (2dp) | A2 | (13 marks)

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7. (a) Express $4 \sin x + 3 \cos x$ in the form $R \sin ( x + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.\\
(b) State the minimum value of $4 \sin x + 3 \cos x$ and the smallest positive value of $x$ for which this minimum value occurs.\\
(c) Solve the equation

$$4 \sin 2 \theta + 3 \cos 2 \theta = 2$$

for $\theta$ in the interval $0 \leq \theta \leq \pi$, giving your answers to 2 decimal places.\\

\hfill \mbox{\textit{Edexcel C3  Q7 [13]}}

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Q1 6 Q2 7 Q3 8 Q4 8 Q5 9 Q6 10 Q7 13 Q8 14

Answer	Marks	Guidance
(a) \(4 \sin x + 3 \cos x = R \sin x \cos \alpha + R \cos x \sin \alpha\)	M1
\(R \cos \alpha = 4\), \(R \sin \alpha = 3\)	M1
\(\therefore R = \sqrt{4^2 + 3^2} = 5\)	M1 A1
\(\tan \alpha = \frac{3}{4}\), \(\alpha = 0.644\) (3sf)	M1 A1
\(\therefore 4 \sin x + 3 \cos x = 5 \sin (x + 0.644)\)	M1
(b) minimum \(= -5\)	B1
occurs when \(x + 0.6435 = \frac{3\pi}{2}\), \(x = 4.07\) (3sf)	M1 A1
(c) \(5 \sin (2\theta + 0.6435) = 2\)	M1
\(\sin (2\theta + 0.6435) = 0.4\)	M1
\(2\theta + 0.6435 = \pi - 0.4115, 2\pi + 0.4115\)	B1 M1
\(2\theta = 2.087, 6.051\)	M1
\(\theta = 1.04, 3.03\) (2dp)	A2	(13 marks)