Question 7 - A-Level Maths

OCR C3 — Question 7 8 marks

Exam Board	OCR
Module	C3 (Core Mathematics 3)
Marks	8
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 three routine parts: (i) converting to R sin(x-α) form using standard formulas, (ii) algebraic manipulation of reciprocal trig functions (multiply by sin x), and (iii) solving using the harmonic form. All techniques are textbook exercises with no novel insight required, making it slightly easier than average.
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

7. (i) Express $2 \sin x ^ { \circ } - 3 \cos x ^ { \circ }$ in the form $R \sin ( x - \alpha ) ^ { \circ }$ where $R > 0$ and $0 < \alpha < 90$.
(ii) Show that the equation $$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2$$ can be written in the form $$2 \sin x ^ { \circ } - 3 \cos x ^ { \circ } = 1$$ (iii) 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

Question 7:

Part (i)

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

Part (ii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\csc x° + 3\cot x° = 2 \Rightarrow \frac{1}{\sin x} + \frac{3\cos x}{\sin x} = 2$
$\Rightarrow 1 + 3\cos x = 2\sin x$
$\Rightarrow 2\sin x° - 3\cos x° = 1$	B1

Part (iii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\sqrt{13}\sin(x - 56.31) = 1$
$\sin(x - 56.31) = \frac{1}{\sqrt{13}}$	M1
$x - 56.31 = 16.10$, $180 - 16.10 = 16.10$, $163.90$	M1
$x = 72.4$, $220.2$ (1dp)	A2	(8)

# Question 7:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R\cos\alpha = 2$, $R\sin\alpha = 3$ | M1 | |
| $R = \sqrt{2^2 + 3^2} = \sqrt{13}$ | A1 | |
| $\tan\alpha = \frac{3}{2}$, $\alpha = 56.3$ (3sf) | A1 | |
| $\therefore 2\sin x° - 3\cos x° = \sqrt{13}\sin(x - 56.3)°$ | | |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\csc x° + 3\cot x° = 2 \Rightarrow \frac{1}{\sin x} + \frac{3\cos x}{\sin x} = 2$ | | |
| $\Rightarrow 1 + 3\cos x = 2\sin x$ | | |
| $\Rightarrow 2\sin x° - 3\cos x° = 1$ | B1 | |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{13}\sin(x - 56.31) = 1$ | | |
| $\sin(x - 56.31) = \frac{1}{\sqrt{13}}$ | M1 | |
| $x - 56.31 = 16.10$, $180 - 16.10 = 16.10$, $163.90$ | M1 | |
| $x = 72.4$, $220.2$ (1dp) | A2 | **(8)** |

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Show LaTeX source

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

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

can be written in the form

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

(iii) 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{OCR C3  Q7 [8]}}

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Q1 5 Q2 6 Q3 7 Q4 7 Q5 8 Q6 8 Q7 8 Q8 10 Q9 13

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\sqrt{13}\sin(x - 56.31) = 1\)
\(\sin(x - 56.31) = \frac{1}{\sqrt{13}}\)	M1
\(x - 56.31 = 16.10\), \(180 - 16.10 = 16.10\), \(163.90\)	M1
\(x = 72.4\), \(220.2\) (1dp)	A2	(8)