Question 4 - A-Level Maths

Edexcel C3 — Question 4 10 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Harmonic Form
Type	Solve reciprocal trig equation
Difficulty	Standard +0.3 This is a standard C3 harmonic form question with clear scaffolding: part (a) is routine R-α conversion, part (b) is given as 'show that' making the algebra straightforward, and part (c) applies the result to solve. While it involves reciprocal trig functions, the question structure guides students through each step, making it slightly easier than average for C3.
Spec	1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc 1.05o Trigonometric equations: solve in given intervals

4. (a) Express $2 \sin x ^ { \circ } - 3 \cos x ^ { \circ }$ in the form $R \sin ( x - \alpha ) ^ { \circ }$ where $R > 0$ and $0 < \alpha < 90$.
(b) Show that the equation $$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2$$ can be written in the form $$2 \sin x ^ { \circ } - 3 \cos x ^ { \circ } = 1 .$$ (c) Solve the equation $$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2 ,$$ for $x$ in the interval $0 \leq x \leq 360$, giving your answers to 1 decimal place.

Show mark scheme Show mark scheme source

Part (a)

Answer	Marks
$2\sin x - 3\cos x = R\sin x \cos \alpha - R\cos x \sin \alpha$	M1 A1
$R\cos \alpha = 2, R\sin \alpha = 3$
$R = \sqrt{2^2 + 3^2} = \sqrt{13}$	M1 A1
$\tan \alpha = \frac{3}{2}, \alpha = 56.3$ (3sf)	M1 A1
$2\sin x° - 3\cos x° = \sqrt{13}\sin(x - 56.3)°$

Part (b)

Answer	Marks
$\operatorname{cosec} x° + 3\cot x° = 2$ ⟹ $\frac{1}{\sin x} + \frac{3\cos x}{\sin x} = 2$	B1
⟹ $1 + 3\cos x = 2\sin x$
⟹ $2\sin^2 x° - 3\cos x° = 1$

Part (c)

Answer	Marks
$\sqrt{13}\sin(x - 56.31) = 1$	M1
$\sin(x - 56.31) = \frac{1}{\sqrt{13}}$
$x - 56.31 = 16.10, 180 - 16.10 = 16.10, 163.90$	B1 M1
$x = 72.4, 220.2$ (1dp)	A2
(10)

**Part (a)**
$2\sin x - 3\cos x = R\sin x \cos \alpha - R\cos x \sin \alpha$ | M1 A1 |
$R\cos \alpha = 2, R\sin \alpha = 3$ |
$R = \sqrt{2^2 + 3^2} = \sqrt{13}$ | M1 A1 |
$\tan \alpha = \frac{3}{2}, \alpha = 56.3$ (3sf) | M1 A1 |
$2\sin x° - 3\cos x° = \sqrt{13}\sin(x - 56.3)°$ |

**Part (b)**
$\operatorname{cosec} x° + 3\cot x° = 2$ ⟹ $\frac{1}{\sin x} + \frac{3\cos x}{\sin x} = 2$ | B1 |
⟹ $1 + 3\cos x = 2\sin x$ |
⟹ $2\sin^2 x° - 3\cos x° = 1$ |

**Part (c)**
$\sqrt{13}\sin(x - 56.31) = 1$ | M1 |
$\sin(x - 56.31) = \frac{1}{\sqrt{13}}$ |
$x - 56.31 = 16.10, 180 - 16.10 = 16.10, 163.90$ | B1 M1 |
$x = 72.4, 220.2$ (1dp) | A2 |
| (10) |

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4. (a) Express $2 \sin x ^ { \circ } - 3 \cos x ^ { \circ }$ in the form $R \sin ( x - \alpha ) ^ { \circ }$ where $R > 0$ and $0 < \alpha < 90$.\\
(b) Show that the equation

$$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2$$

can be written in the form

$$2 \sin x ^ { \circ } - 3 \cos x ^ { \circ } = 1 .$$

(c) Solve the equation

$$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2 ,$$

for $x$ in the interval $0 \leq x \leq 360$, giving your answers to 1 decimal place.\\

\hfill \mbox{\textit{Edexcel C3  Q4 [10]}}

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Q1 7 Q2 9 Q3 10 Q4 10 Q5 12 Q6 13 Q7 14

Answer	Marks
\(2\sin x - 3\cos x = R\sin x \cos \alpha - R\cos x \sin \alpha\)	M1 A1
\(R\cos \alpha = 2, R\sin \alpha = 3\)
\(R = \sqrt{2^2 + 3^2} = \sqrt{13}\)	M1 A1
\(\tan \alpha = \frac{3}{2}, \alpha = 56.3\) (3sf)	M1 A1
\(2\sin x° - 3\cos x° = \sqrt{13}\sin(x - 56.3)°\)

Answer	Marks
\(\operatorname{cosec} x° + 3\cot x° = 2\) ⟹ \(\frac{1}{\sin x} + \frac{3\cos x}{\sin x} = 2\)	B1
⟹ \(1 + 3\cos x = 2\sin x\)
⟹ \(2\sin^2 x° - 3\cos x° = 1\)

Answer	Marks
\(\sqrt{13}\sin(x - 56.31) = 1\)	M1
\(\sin(x - 56.31) = \frac{1}{\sqrt{13}}\)
\(x - 56.31 = 16.10, 180 - 16.10 = 16.10, 163.90\)	B1 M1
\(x = 72.4, 220.2\) (1dp)	A2
(10)