Question 6 - A-Level Maths

OCR C3 — Question 6 10 marks

Exam Board	OCR
Module	C3 (Core Mathematics 3)
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Harmonic Form
Type	Express and solve equation
Difficulty	Standard +0.3 This is a standard harmonic form question with routine steps: finding R and α using Pythagorean identity and tan, stating maximum from amplitude, and solving a transformed equation. All techniques are textbook exercises for C3 level 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

6. (i) Express $\sqrt { 3 } \sin \theta + \cos \theta$ in the form $R \sin ( \theta + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.
(ii) State the maximum value of $\sqrt { 3 } \sin \theta + \cos \theta$ and the smallest positive value of $\theta$ for which this maximum value occurs.
(iii) 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 $\pi$.

Show mark scheme Show mark scheme source

Question 6:

Part (i)

Answer	Marks	Guidance
Answer	Mark	Notes
$R\cos\alpha = \sqrt{3}$, $R\sin\alpha = 1$	M1
$R = \sqrt{3+1} = 2$	A1
$\tan\alpha = \frac{1}{\sqrt{3}}$, $\alpha = \frac{\pi}{6}$	A1
$\therefore \sqrt{3}\sin\theta + \cos\theta = 2\sin\left(\theta + \frac{\pi}{6}\right)$

Part (ii)

Answer	Marks	Guidance
Answer	Mark	Notes
maximum $= 2$	B1
occurs when $\theta + \frac{\pi}{6} = \frac{\pi}{2}$, $\theta = \frac{\pi}{3}$	M1 A1

Part (iii)

Answer	Marks	Guidance
Answer	Mark	Notes
$\sin\left(\theta + \frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$	M1
$\theta + \frac{\pi}{6} = -\frac{\pi}{3}$, $-\pi + \frac{\pi}{3} = -\frac{\pi}{3}$, $-\frac{2\pi}{3}$	M1
$\theta = -\frac{5\pi}{6}$, $-\frac{\pi}{2}$	A2	(10)

# Question 6:

## Part (i)
| Answer | Mark | Notes |
|--------|------|-------|
| $R\cos\alpha = \sqrt{3}$, $R\sin\alpha = 1$ | M1 | |
| $R = \sqrt{3+1} = 2$ | A1 | |
| $\tan\alpha = \frac{1}{\sqrt{3}}$, $\alpha = \frac{\pi}{6}$ | A1 | |
| $\therefore \sqrt{3}\sin\theta + \cos\theta = 2\sin\left(\theta + \frac{\pi}{6}\right)$ | | |

## Part (ii)
| Answer | Mark | Notes |
|--------|------|-------|
| maximum $= 2$ | B1 | |
| occurs when $\theta + \frac{\pi}{6} = \frac{\pi}{2}$, $\theta = \frac{\pi}{3}$ | M1 A1 | |

## Part (iii)
| Answer | Mark | Notes |
|--------|------|-------|
| $\sin\left(\theta + \frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$ | M1 | |
| $\theta + \frac{\pi}{6} = -\frac{\pi}{3}$, $-\pi + \frac{\pi}{3} = -\frac{\pi}{3}$, $-\frac{2\pi}{3}$ | M1 | |
| $\theta = -\frac{5\pi}{6}$, $-\frac{\pi}{2}$ | A2 | **(10)** |

---

Show LaTeX source

6. (i) Express $\sqrt { 3 } \sin \theta + \cos \theta$ in the form $R \sin ( \theta + \alpha )$ where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.\\
(ii) State the maximum value of $\sqrt { 3 } \sin \theta + \cos \theta$ and the smallest positive value of $\theta$ for which this maximum value occurs.\\
(iii) 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{OCR C3  Q6 [10]}}

This paper (8 questions)

View full paper

Q1 5 Q2 6 Q3 8 Q4 9 Q5 10 Q6 10 Q7 10 Q8 14

Answer	Marks	Guidance
Answer	Mark	Notes
\(R\cos\alpha = \sqrt{3}\), \(R\sin\alpha = 1\)	M1
\(R = \sqrt{3+1} = 2\)	A1
\(\tan\alpha = \frac{1}{\sqrt{3}}\), \(\alpha = \frac{\pi}{6}\)	A1
\(\therefore \sqrt{3}\sin\theta + \cos\theta = 2\sin\left(\theta + \frac{\pi}{6}\right)\)

Answer	Marks	Guidance
Answer	Mark	Notes
\(\sin\left(\theta + \frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}\)	M1
\(\theta + \frac{\pi}{6} = -\frac{\pi}{3}\), \(-\pi + \frac{\pi}{3} = -\frac{\pi}{3}\), \(-\frac{2\pi}{3}\)	M1
\(\theta = -\frac{5\pi}{6}\), \(-\frac{\pi}{2}\)	A2	(10)