November 2021 Paper 2 Q20
20 Show that \(\quad \dfrac{2\sin x + \cos x}{\tan x} - \dfrac{1}{\sin x} \quad\) can be written in the form \(\quad a\cos x + b\sin x\)
where \(a\) and \(b\) are integers. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Uses a common denominator of \(\sin x\) | ||
| \(\dfrac{2\sin x + \cos x}{\dfrac{\sin x}{\cos x}} - \dfrac{1}{\sin x}\) or \(\dfrac{\cos x(2\sin x + \cos x)}{\sin x} - \dfrac{1}{\sin x}\) | M1 | implied by 2nd M1 condone omission of \(-\dfrac{1}{\sin x}\) |
| \(\dfrac{2\sin x\cos x + \cos^2 x}{\sin x} - \dfrac{1}{\sin x}\) or \(2\cos x + \dfrac{\cos^2 x}{\sin x} - \dfrac{1}{\sin x}\) | M1dep | condone omission of \(-\dfrac{1}{\sin x}\) |
| \(\dfrac{2\sin x\cos x + \cos^2 x - \cos^2 x - \sin^2 x}{\sin x}\) or \(\dfrac{2\sin x\cos x + 1 - \sin^2 x - 1}{\sin x}\) or \(\dfrac{2\sin x\cos x - \sin^2 x}{\sin x}\) or \(2\cos x - \dfrac{\sin^2 x}{\sin x}\) | M1dep | |
| \(2\cos x - \sin x\) with M3 | A1 | |
| Alternative method 2 Uses a common denominator of \(\tan x\sin x\) | ||
| \(\dfrac{\sin x(2\sin x + \cos x) - \tan x}{\tan x\sin x}\) or \(\dfrac{2\sin^2 x + \cos x\sin x - \tan x}{\tan x\sin x}\) | M1 | \(\tan x\) may be seen as \(\dfrac{\sin x}{\cos x}\) |
| \(\dfrac{2\sin^2 x\cos x + \cos^2 x\sin x - \sin x}{\sin^2 x}\) | M1dep | |
| \(\dfrac{2\sin x\cos x + \cos^2 x - \cos^2 x - \sin^2 x}{\sin x}\) or \(\dfrac{2\sin x\cos x + 1 - \sin^2 x - 1}{\sin x}\) or \(\dfrac{2\sin x\cos x - \sin^2 x}{\sin x}\) | M1dep | allow the fractions with denominator \(\sin^2 x\) eg \(\dfrac{2\sin^2 x\cos x + \sin x - \sin^3 x - \sin x}{\sin^2 x}\) |
| \(2\cos x - \sin x\) with M3 | A1 | |
Additional guidance
| \(2\cos x - \sin x\) with no method | Zero |
| Allow eg \(1\sin x\) for \(\sin x\) | |
| Condone \(x\) missing for M marks and award A mark if recovered | |
| Fractions may be seen as separate fractions/terms eg1 Alt 1 \(\dfrac{2\sin x}{\dfrac{\sin x}{\cos x}} + \dfrac{\cos x}{\dfrac{\sin x}{\cos x}} - \dfrac{1}{\sin x}\) \(2\cos x + \dfrac{\cos^2 x}{\sin x} - \dfrac{1}{\sin x}\) \(2\cos x + \dfrac{1}{\sin x} - \dfrac{\sin^2 x}{\sin x} - \dfrac{1}{\sin x}\) eg2 Alt 2 \(\dfrac{2\sin^2 x + \cos x\sin x}{\tan x\sin x} - \dfrac{\tan x}{\tan x\sin x}\) \(\dfrac{2\sin^2 x\cos x + \cos^2 x\sin x}{\sin^2 x} - \dfrac{\sin x}{\sin^2 x}\) \(\dfrac{2\sin x\cos x + \cos^2 x}{\sin x} - \dfrac{\cos^2 x + \sin^2 x}{\sin x}\) | M1 M1 M1 M1 M1 M1 |