June 2023 Paper 2 Q20
20 Show that
\[2\sin^3 x + 2\sin x\cos^2 x + 5\tan x\cos x \quad \text{simplifies to} \quad p\sin x \quad \text{where } p \text{ is a constant.}\][3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2\sin x(\sin^2 x + \cos^2 x)\) or \(\sin x(2\sin^2 x + 2\cos^2 x)\) or \(2\sin x(1 - \sin^2 x)\) or \(2\sin x(1 - \cos^2 x)\) | M1 | oe writing first two terms using \(\sin^2 x + \cos^2 x\) or \(2\sin^2 x + 2\cos^2 x\) or replacing \(\cos^2 x\) with \((1 - \sin^2 x)\) in 2nd term or replacing \(\sin^2 x\) with \((1 - \cos^2 x)\) in 1st term ignore other terms |
| \(5\dfrac{\sin x}{\cos x}\,(\times)\cos x\) or \(5\sin x\) | M1 | oe replacing \(\tan x\) with \(\dfrac{\sin x}{\cos x}\) in 3rd term ignore other terms |
| \(7\sin x\) with M2 seen | A1 | allow \(p = 7\) with M2 seen |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| M marks are independent and can be scored in either order eg \(2\sin^3 x + 2\sin x\cos^2 x + 5\sin x\) \(\sin x(2\sin^2 x + 2\cos^2 x + 5)\) \(7\sin x\) | M1 (2nd) M1 (1st) A1 |
| Allow eg \(7 \times \sin x\) or \(\sin x \times 7\) for \(7\sin x\) | |
| Missing brackets must be recovered | |
| NOTE \(\;7\sin x\) may come from incorrect or incomplete working eg \(2\sin^2 x + 2\cos^2 x + 5\sin x = 7\sin x\) | M0M1A0 |
| Condone unambiguous notation eg1 condone s or sin for \(\sin x\) eg2 condone \(\cos\theta\) for \(\cos x\) |