June 2019 Paper 2 Q22
22
(a) Show that \(\quad \dfrac{2\sin^2 x - 1 + \cos^2 x}{\sin x \cos x} \quad\) is equivalent to \(\ \tan x\) [3 marks]
(b) Hence solve \(\quad \dfrac{2\sin^2 x - 1 + \cos^2 x}{\sin x \cos x} = -1 \quad\) for \(\ 0^\circ \leqslant x \leqslant 360^\circ\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(2\sin^2 x - 1 + 1 - \sin^2 x\) or \(2\sin^2 x - (\sin^2 x + \cos^2 x) + \cos^2 x\) or \(2\sin^2 x - \sin^2 x - \cos^2 x + \cos^2 x\) or \(2\sin^2 x - \sin^2 x\) or \(\sin^2 x - \cos^2 x + \cos^2 x\) or \(1 + \sin^2 x - 1\) | M1 | use of \(\sin^2 x + \cos^2 x = 1\) in numerator ignore any denominator |
| \(\dfrac{\sin^2 x}{\sin x \cos x}\) with M1 seen or \(\dfrac{\sin^2 x}{\tan x \cos^2 x}\) with M1 seen | M1dep | simplification to one step from \(\dfrac{\sin x}{\cos x}\) or simplification to one step from \(\dfrac{\tan^2 x}{\tan x}\) |
| \(\dfrac{\sin x}{\cos x}\) and \(\tan x\) with M2 seen or \(\dfrac{\tan^2 x}{\tan x}\) and \(\tan x\) with M2 seen | A1 | SC3 equates given expression to \(\tan x\) and cross multiplies to show equivalence with full working shown |
| Alternative method 2 | ||
| \(2(1 - \cos^2 x) - 1 + \cos^2 x\) or \(2 - 2\cos^2 x - 1 + \cos^2 x\) | M1 | use of \(\sin^2 x + \cos^2 x = 1\) in numerator ignore any denominator |
| \(\dfrac{1 - \cos^2 x}{\sin x \cos x}\) and \(\dfrac{\sin^2 x}{\sin x \cos x}\) with M1 seen or \(\dfrac{1 - \cos^2 x}{\sin x \cos x}\) and \(\dfrac{\sin^2 x}{\tan x \cos^2 x}\) with M1 seen | M1dep | simplification to one step from \(\dfrac{\sin x}{\cos x}\) or simplification to one step from \(\dfrac{\tan^2 x}{\tan x}\) |
| \(\dfrac{\sin x}{\cos x}\) and \(\tan x\) with M2 seen or \(\dfrac{\tan^2 x}{\tan x}\) and \(\tan x\) with M2 seen | A1 | SC3 equates given expression to \(\tan x\) and cross multiplies to show equivalence with full working shown |
| Alternative method 3 | ||
| \(\dfrac{2\sin x}{\cos x} - \dfrac{\sin^2 x}{\sin x \cos x}\) | M1 | from \(\dfrac{2\sin^2 x}{\sin x \cos x} - \dfrac{1 - \cos^2 x}{\sin x \cos x}\) |
| \(2\tan x - \dfrac{\sin^2 x}{\sin x \cos x}\) or \(\dfrac{2\sin x}{\cos x} - \dfrac{\sin x}{\cos x}\) with M1 seen | M1dep | simplification to one step from \(2\tan x - \tan x\) |
| \(2\tan x - \tan x\) and \(\tan x\) with M2 seen | A1 | SC3 equates given expression to \(\tan x\) and cross multiplies to show equivalence with full working shown |
Additional guidance
| Equating given expression to \(\tan x\) and cross multiplying can score SC3 or M1M0A0 eg1 Alt 1 \(\dfrac{2\sin^2 x - 1 + \cos^2 x}{\sin x \cos x} = \tan x\) \(2\sin^2 x - 1 + \cos^2 x = \tan x \sin x \cos x\) \(2\sin^2 x - 1 + 1 - \sin^2 x = \tan x \sin x \cos x \quad\) (scores M1 here for LHS) eg2 \(\dfrac{2\sin^2 x - 1 + \cos^2 x}{\sin x \cos x} = \tan x\) \(2\sin^2 x - 1 + \cos^2 x = \tan x \sin x \cos x\) \(2\sin^2 x - 1 + 1 - \sin^2 x = \tan x \sin x \cos x\) \(\sin^2 x = \tan x \sin x \cos x\) \(\sin^2 x = \dfrac{\sin x}{\cos x} \sin x \cos x\) \(\sin^2 x = \sin^2 x\) | M1M0A0 SC3 |
| Use of \(\sin x = \dfrac{\text{opp}}{\text{hyp}}\) etc | M0M0A0 |
| Allow sin or s for \(\sin x\) etc | |
| Condone \(\sin x^2\) for \(\sin^2 x\) etc | |
| Allow any letter for \(x\) | |
| Alts 1 and 2 For A1 \(\dfrac{\sin x}{\cos x}\) is implied by \(\dfrac{\sin^2 x}{\sin x \cos x}\) with cancelling shown |
| Answer | Mark | Comments |
|---|---|---|
| 135 and 315 with no other solutions [0, 360] | B2 | B1 135 with no other solutions [0, 360] or 315 with no other solutions [0, 360] SC1 135 and 315 with one other solution [0, 360] |
Additional guidance
| Mark the answer line unless blank eg 135 and 315 in working with 135 on answer line | B1 |
| \(-45\) and 135 and 315 | B2 |
| \(-45\) and 135 | B1 |
| Ignore incorrect solutions outside the range [0, 360] eg 135 and 315 and \(-90\) | B2 |
| 135 and 225 and 315 | SC1 |
| Both answers embedded ie tan 135 tan 315 | B1 |
| 0 and 135 and 225 and 315 | B0 |
| 45 and 135 | B0 |
| 225 and 315 | B0 |