AS June 2024 Q4
4.
| Scheme | Marks | AO |
|---|---|---|
| \(\cos x = \cos^2\dfrac{x}{2} - \sin^2\dfrac{x}{2}\) or \(2\cos^2\dfrac{x}{2} - 1\) or \(1 - 2\sin^2\dfrac{x}{2}\) or \(\tan x = \dfrac{\tan\frac{x}{2} + \tan\frac{x}{2}}{1 - \tan^2\frac{x}{2}}\) | B1 | 1.2 |
| \(\Rightarrow \cos x = \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2 - \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2\) or \(2\left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2 - 1\) or \(1 - 2\left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2\) or \(\tan x = \dfrac{2t}{1 - t^2} \Rightarrow \cos x = \dfrac{1 - t^2}{\sqrt{(1 - t^2)^2 + 4t^2}}\) | M1 | 1.1b |
| \(= \dfrac{1}{1 + t^2} - \dfrac{t^2}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2}\ *\) or \(= \dfrac{2}{1 + t^2} - 1 = \dfrac{2 - 1 - t^2}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2}\ *\) or \(1 - \dfrac{2t^2}{1 + t^2} = \dfrac{1 + t^2 - 2t^2}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2}\ *\) or \(\dfrac{1 - t^2}{\sqrt{(1 - t^2)^2 + 4t^2}} = \dfrac{1 - t^2}{\sqrt{t^4 + 2t^2 + 1}} = \dfrac{1 - t^2}{\sqrt{(1 + t^2)^2}} = \dfrac{1 - t^2}{1 + t^2}\ *\) | A1* | 2.1 |
| (3) |
Notes
B1: Uses any correct appropriate double angle identity for \(\cos x\) or possibly \(\tan x\).
M1: Substitutes correct expressions for \(\cos\dfrac{x}{2}\) and/or \(\sin\dfrac{x}{2}\) in terms of \(t\)
A1*: Completes the proof with sufficient working shown with no errors.
Other alternatives are possible and can be marked in a similar way e.g.
\(\cos x = \dfrac{\sin x}{\tan x} = \dfrac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\left(\dfrac{2\tan\frac{x}{2}}{1 - \tan^2\frac{x}{2}}\right)}\)
Scores B1 for a correct double angle identity for \(\cos x\)
\(= 2\left(\dfrac{1}{\sqrt{1 + t^2}}\right)\left(\dfrac{t}{\sqrt{1 + t^2}}\right) \times \dfrac{1 - t^2}{2t}\)
M1 for substituting correct expressions for \(\cos\dfrac{x}{2}\), \(\sin\dfrac{x}{2}\) and \(\tan\dfrac{x}{2}\) in terms of \(t\)
\(= \dfrac{2t}{1 + t^2} \times \dfrac{1 - t^2}{2t} = \dfrac{1 - t^2}{1 + t^2}\ *\)
A1* for completing the proof with sufficient working shown with no errors.
Allow \(\theta\) for \(x\) for the first 2 marks but must obtain \(\cos x = \ldots\) for A1*
(a) Special Case – candidates who quote results for tan x and/or sin x to verify the result:
e.g.
\(t = \tan\dfrac{x}{2} \Rightarrow \tan x = \dfrac{2t}{1 - t^2},\ \sin x = \dfrac{2t}{1 + t^2}\)
\(\cos x = \dfrac{\sin x}{\tan x} = \dfrac{2t}{1 + t^2} \times \dfrac{1 - t^2}{2t} = \dfrac{1 - t^2}{1 + t^2}\)
or e.g.
\(t = \tan\dfrac{x}{2} \Rightarrow \sin x = \dfrac{2t}{1 + t^2}\)
\(\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \dfrac{4t^2}{(1 + t^2)^2}} = \sqrt{\dfrac{1 + 2t^2 + t^4 - 4t^2}{(1 + t^2)^2}} = \sqrt{\dfrac{(1 - t^2)^2}{(1 + t^2)^2}} = \dfrac{1 - t^2}{1 + t^2}\)
Scores SC B1 only
| Scheme | Marks | AO |
|---|---|---|
| \(3\tan x - 10\cos x = 10 \Rightarrow 3 \times \dfrac{2t}{1 - t^2} - 10 \times \dfrac{1 - t^2}{1 + t^2} = 10\) | B1 | 1.1a |
| \(\Rightarrow 6t(1 + t^2) - 10(1 - t^2)^2 = 10(1 - t^2)(1 + t^2)\) \(\Rightarrow 6t + 6t^3 - 10 + 20t^2 - 10t^4 = 10 - 10t^4\) \(\Rightarrow 6t^3 + 20t^2 + 6t - 20 = 0\) or e.g. \(3t^3 + 10t^2 + 3t - 10 = 0\) | M1 | 2.1 |
| \(\Rightarrow (t + 2)(\text{“}6\text{”}t^2 + \ldots t + \ldots) = 0\) or e.g. \(\Rightarrow (t + 2)(\text{“}3\text{”}t^2 + \ldots t + \ldots) = 0\) | M1 | 1.1b |
| \(\Rightarrow (t + 2)(6t^2 + 8t - 10) = 0\) or \((t + 2)(3t^2 + 4t - 5) = 0\) | A1 | 2.2a |
| (4) |
Notes
B1: Applies part (a) and the half angle formula for tan to give a correct equation in \(t\).
M1: Attempts to multiply by \(1 - t^2\) and \(1 + t^2\), expands and simplifies to a cubic in \(t\).
M1: Attempts to take a factor \((t + 2)\) out of their cubic.
May be done by inspection or long division but must obtain the correct coefficient for \(t^2\) and reach a 3 term quadratic expression.
A1: Correct equation achieved as shown. Accept with any integer multiple of the quadratic.
Must see the equation written down not just values for \(a\), \(b\) and \(c\).
| Scheme | Marks | AO |
|---|---|---|
| \(t + 2 = 0 \Rightarrow x = 2 \times \arctan(-2) = \ldots\) | M1 | 1.1b |
| \(x = -126.86\ldots^\circ\) (allow awrt \(-127^\circ\)) | A1 | 2.2a |
| \(3t^2 + 4t - 5 = 0 \Rightarrow t = \dfrac{-4 \pm \sqrt{16 - 4 \times 3 \times -5}}{6} = \dfrac{-2 \pm \sqrt{19}}{3}\) \((= 0.786\ldots, -2.11\ldots)\) \(\Rightarrow x = 2 \times \arctan(\ldots) = \ldots\) | M1 | 3.1a |
| \(x = \text{(awrt) } -129^\circ, 76.4^\circ \quad (-129.486\ldots, 76.355\ldots)\) | A1 (one) A1 (both) | 1.1b 1.1b |
| (5) | ||
| (12 marks) |
Notes
M1: Attempts to solve the \(t + 2 = 0\) equation, look for an attempt at \(\arctan(\pm 2)\) and an attempt to double. Allow in radians for the M mark.
A1: awrt \(-127^\circ\)
M1: Solves their quadratic and attempts arctan and doubles.
The usual rules apply for solving the quadratic and may be implied by their values.
A1: Either awrt \(-129^\circ\) or awrt \(76.4^\circ\) (degrees symbol not required) or allow for this mark one of these in radians (awrt \(-2.25\) or awrt 1.33)
A1: Both awrt \(-129^\circ\) (or better e.g. \(-129.5^\circ\)) and awrt \(76.4^\circ\) (degrees symbol not required)
There must be no other values in range for this mark. Values outside the range can be ignored.