June 2025 Paper 3 Q13
13
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
The relevant parts of the article “The trisectrix of Maclaurin” are reproduced below; the line numbers are those printed on the Insert.
Lines 24–25
Using the formula for \(\tan(A+B)\) in terms of \(\tan A\) and \(\tan B\), it can be shown that
\(\tan 3\theta = \dfrac{t(3-t^2)}{1-3t^2}\), where \(t = \tan\theta\).
Use \(\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\), with suitable choices for \(A\) and \(B\), to show that \(\tan 3\theta = \dfrac{t(3-t^2)}{1-3t^2}\), where \(t = \tan\theta\), as given in line 25. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\tan(2\theta) = \frac{2t}{1-t^2}\) | B1 | 1.1 |
| \((\tan(3\theta) = \tan(2\theta + \theta) =)\ \dfrac{\frac{2t}{1-t^2} + t}{1 - \frac{2t^2}{1-t^2}}\) | M1 | 3.1a |
| \(\tan(3\theta) = \dfrac{2t + t(1-t^2)}{1 - t^2 - 2t^2}\) \(\qquad = \dfrac{3t - t^3}{1 - 3t^2}\) \(\qquad = \dfrac{t(3-t^2)}{1-3t^2}\) | A1 | 2.1 |
| [3] |
Notes
B1: OR \(\tan(3\theta) = \tan(2\theta + \theta) = \frac{\tan 2\theta + \tan\theta}{1 - \tan 2\theta\tan\theta}\)
Formula for \(\tan(2\theta)\) in terms of \(t\) or \(\tan\theta\)
Allow embedded
M1: Must be in terms of \(\tan\theta\) or \(t\) only
Condone considering numerator and denominator separately
Using their \(\tan(2\theta)\) provided from using double angle formula
A1: AG
Correct completion to given result in terms of \(t\).
Must see at least one step of working after M1 awarded and no errors or omissions