June 2016 Paper 2 Q26
26
(a) Prove that \(\quad \sin^2 x - 3\cos^2 x \equiv 4\sin^2 x - 3\) [2 marks]
(b) Hence, or otherwise, work out the values of \(x\) between 0° and 360° for which\[\sin^2 x - 3\cos^2 x = 0\] [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 (LHS \(\to\) RHS) | ||
| \(\sin^2 x - 3(1 - \sin^2 x)\) | M1 | Must see \((1 - \sin^2 x)\) |
| \(\sin^2 x - 3 + 3\sin^2 x = 4\sin^2 x - 3\) | A1 | Must see correct expansion SC1 Correct rearrangement of given identity to \(3\sin^2 x + 3\cos^2 x = 3\) and \(3(\sin^2 x + \cos^2 x) = 3\) and \(\sin^2 x + \cos^2 x = 1\) |
| Alternative method 2 (LHS \(\to\) RHS) | ||
| \(1 - \cos^2 x - 3\cos^2 x = 1 - 4\cos^2 x\) \(= 1 - 4(1 - \sin^2 x)\) | M1 | Must see \((1 - \cos^2 x)\) and \((1 - \sin^2 x)\) |
| \(1 - 4 + 4\sin^2 x = 4\sin^2 x - 3\) | A1 | Must see correct expansion SC1 Correct rearrangement of given identity to \(3\sin^2 x + 3\cos^2 x = 3\) and \(3(\sin^2 x + \cos^2 x) = 3\) and \(\sin^2 x + \cos^2 x = 1\) |
| Alternative method 3 (RHS \(\to\) LHS) | ||
| \(4\sin^2 x - 3(\sin^2 x + \cos^2 x)\) | M1 | Must see \((\sin^2 x + \cos^2 x)\) |
| \(4\sin^2 x - 3\sin^2 x - 3\cos^2 x\) \(= \sin^2 x - 3\cos^2 x\) | A1 | Must see correct expansion SC1 Correct rearrangement of given identity to \(3\sin^2 x + 3\cos^2 x = 3\) and \(3(\sin^2 x + \cos^2 x) = 3\) and \(\sin^2 x + \cos^2 x = 1\) |
| Alternative method 4 (RHS \(\to\) LHS) | ||
| \(4(1 - \cos^2 x) - 3 = 4 - 4\cos^2 x - 3\) \(= 1 - 4\cos^2 x\) \(= \sin^2 x + \cos^2 x - 4\cos^2 x\) | M1 | Must see \((1 - \cos^2 x)\) and \(\sin^2 x + \cos^2 x\) and correct expansion |
| \(= \sin^2 x - 3\cos^2 x\) | A1 | SC1 Correct rearrangement of given identity to \(3\sin^2 x + 3\cos^2 x = 3\) and \(3(\sin^2 x + \cos^2 x) = 3\) and \(\sin^2 x + \cos^2 x = 1\) |
| Alternative method 5 (LHS and RHS \(\to\) common expression) | ||
| \(1 - \cos^2 x - 3\cos^2 x = 1 - 4\cos^2 x\) and \(4(1 - \cos^2 x) - 3 = 4 - 4\cos^2 x - 3\) \(= 1 - 4\cos^2 x\) | B2 | Must see \((1 - \cos^2 x)\) and correct expansion SC1 Correct rearrangement of given identity to \(3\sin^2 x + 3\cos^2 x = 3\) and \(3(\sin^2 x + \cos^2 x) = 3\) and \(\sin^2 x + \cos^2 x = 1\) |
Additional guidance
| As shown in the mark scheme, allow = signs but they may be seen (correctly) as the identity symbol | |
| = signs may be implied (eg working down the page, line by line) | |
| To give M1 the working must not need any further identities applying | |
| The other side of the identity may be seen throughout working in Alts 1 to 4 However, full working on one side of the identity is needed for M1 A1 eg (Alt 2) \(1 - \cos^2 x - 3\cos^2 x = 4\sin^2 x - 3\) \(1 - 4\cos^2 x = 4\sin^2 x - 3\) \(1 - 4(1 - \sin^2 x) = 4\sin^2 x - 3\) \(1 - 4 + 4\sin^2 x = 4\sin^2 x - 3\) (with \(4\sin^2 x - 3 = 4\sin^2 x - 3\) it would be M1 A1) | M1 A0 |
| Other examples may be seen, escalate if necessary | |
| Allow any variable or mixed variables or no variables | |
| Allow \((\sin x)^2\) for \(\sin^2 x\) and \((\cos x)^2\) for \(\cos^2 x\) Allow \(\mathrm{s}^2\) for \(\sin^2 x\) and \(\mathrm{c}^2\) for \(\cos^2 x\) | |
| Do not allow \(\sin x^2\) for \(\sin^2 x\) (but could still gain M1) eg1 Alt 1 \(\sin^2 x - 3(1 - \sin^2 x)\) \(= \sin^2 x - 3 + 3\sin x^2 = 4\sin x^2 - 3\) eg1 Alt 1 \(\sin x^2 - 3(1 - \sin^2 x)\) \(= \sin^2 x - 3 + 3\sin x^2 = 4\sin x^2 - 3\) | M1 A0 M0 A0 |
| Do not allow recovery of missing brackets as this is a proof | |
| SC1 Instead of factorisation, they can divide by 3 | |
| Other examples of SC1 may be seen where the identity is assumed to be correct and correct working with use of \(\sin^2 x + \cos^2 x = 1\) is seen |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin^2 x = \dfrac{3}{4}\) or \(\sin x = \dfrac{\sqrt{3}}{2}\) or \(\sin x = \sqrt{\dfrac{3}{4}}\) or 60 or 120 | M1 | oe eg \((\sin x)^2 = \dfrac{3}{4}\) Allow 0.86... or 0.87 for \(\dfrac{\sqrt{3}}{2}\) Must have \(\sin^2 x =\) or \(\sin x =\) or \(\sin^{-1}\) Allow s for \(\sin x\) Do not allow \(\sin x^2\) for \(\sin^2 x\) but may be recovered |
| \(\sin x = -\dfrac{\sqrt{3}}{2}\) or \(\sin x = -\sqrt{\dfrac{3}{4}}\) or 240 or 300 or \(-60\) | M1 | oe Allow \(-0.86\)... or \(-0.87\) for \(-\dfrac{\sqrt{3}}{2}\) |
| 60 and 120 and 240 and 300 with no other angles in range | A2 | A1 60 and 120 or 240 and 300 |
| Alternative method 2 | ||
| \(\tan^2 x = 3\) or \(\tan x = \sqrt{3}\) or 60 or 240 | M1 | oe eg \((\tan x)^2 = 3\) Allow 1.73... for \(\sqrt{3}\) Must have \(\tan^2 x =\) or \(\tan x =\) or \(\tan^{-1}\) Allow t for \(\tan x\) Do not allow \(\tan x^2\) for \(\tan^2 x\) but may be recovered |
| \(\tan x = -\sqrt{3}\) or 120 or 300 or \(-60\) | M1 | Allow \(-1.73\)... for \(-\sqrt{3}\) |
| 60 and 120 and 240 and 300 with no other angles in range | A2 | A1 60 and 240 or 120 and 300 |
| Alternative method 3 | ||
| \(\cos^2 x = \dfrac{1}{4}\) or \(\cos x = \dfrac{1}{2}\) or \(\cos x = \sqrt{\dfrac{1}{4}}\) or 60 or 300 | M1 | oe eg \((\cos x)^2 = \dfrac{1}{4}\) Must have \(\cos^2 x =\) or \(\cos x =\) or \(\cos^{-1}\) Allow c for \(\cos x\) Do not allow \(\cos x^2\) for \(\cos^2 x\) but may be recovered |
| \(\cos x = -\dfrac{1}{2}\) or \(\cos x = -\sqrt{\dfrac{1}{4}}\) or 120 or 240 | M1 | oe |
| 60 and 120 and 240 and 300 with no other angles in range | A2 | A1 60 and 300 or 120 and 240 |
Additional guidance
| Ignore any solutions outside of \(0 \lt x \lt 360\) ie 0 and 360 are outside the range and can be ignored | |
| All four solutions with extra solutions in range, \(0 \lt x \lt 360\), are penalised one accuracy mark eg 60 90 120 150 240 300 Only penalise extra solutions in range when all four correct solutions are given | M1 M1 A1 |
| Answer line blank, award any marks gained from working lines | |
| If angles are found in working lines but only some are listed on answer line award any method marks gained from the working lines award any accuracy marks gained from the answer line eg1 Working lines \(\sin x = \pm\dfrac{\sqrt{3}}{2}\) 60 and 120 and 240 and 300 Answer line 60 and 120 and 240 eg2 Working lines \(\tan x = \sqrt{3}\) 60 240 Answer line 60 eg3 Working lines \(\sin x = \dfrac{\sqrt{3}}{2}\) 60 120 \(\sin x = -\dfrac{\sqrt{3}}{2}\) 300 Answer line 300 | M1 M1 A1 M1 M0 A0 M1 M1 A0 |
| Answers only can score up to 4 marks All 4 correct \(\to\) 4 marks 3 correct \(\to\) 3 marks 2 correct \(\to\) 2 marks 1 correct \(\to\) 1 mark | |
| M1 M0 A1 or M0 M1 A1 are possible eg1 \(\sin x = \dfrac{\sqrt{3}}{2}\) 60 120 eg2 \(\sin x = -\dfrac{\sqrt{3}}{2}\) 240 300 | M1 M0 A1 M0 M1 A1 |
| Embedded answers can score up to M1 M1 A0 | |
| Working in rads or grads can score M marks if method seen |