A2 June 2025 Paper 1 Q10
10 In this question you must show detailed reasoning.
\(\tan 5\theta \equiv \dfrac{\tan^5\theta - 10\tan^3\theta + 5\tan\theta}{5\tan^4\theta - 10\tan^2\theta + 1}\). [4]
\(t^4 - 4t^3 - 14t^2 - 4t + 1 = 0\).
Give the roots in the form \(t = \tan\phi\) where \(0 \lt \phi \lt \pi\). [4]
\(\tan\left(\dfrac{9}{20}\pi\right) = 1 + \sqrt{5} + \sqrt{5 + 2\sqrt{5}}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\cos 5\theta + \mathrm{i}\sin 5\theta = (c + \mathrm{i}s)^5\) | M1* | 1.1 |
| \(= c^5 + 5\mathrm{i}c^4s - 10c^3s^2 - 10\mathrm{i}c^2s^3 + 5cs^4 + \mathrm{i}s^5\) | M1dep* | 1.1 |
| \(\tan 5\theta \left(= \dfrac{\mathrm{Im}(c + \mathrm{i}s)^5}{\mathrm{Re}(c + \mathrm{i}s)^5}\right) = \dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4}\) | M1 | 1.1 |
| \(\tan 5\theta = \dfrac{\dfrac{5c^4s}{c^5} - \dfrac{10c^2s^3}{c^5} + \dfrac{s^5}{c^5}}{\dfrac{c^5}{c^5} - \dfrac{10c^3s^2}{c^5} + \dfrac{5cs^4}{c^5}}\) \(= \dfrac{\tan^5\theta - 10\tan^3\theta + 5\tan\theta}{5\tan^4\theta - 10\tan^2\theta + 1}\) | A1 | 2.1 |
| [4] |
Notes
M1*: De Moivre’s theorem with \(n = 5\)
Assume \(c = \cos\theta\) and \(s = \sin\theta\) (so condone not explicitly stated)
M1dep*: Expanding \((c + \mathrm{i}s)^5\) to obtain six terms - coefficients must be numerical and correct (so binomial coefficients must be evaluated) and correctly in terms of i only (so not powers of i). Allow at most one sign error and at most one term with an incorrect index
M1: Correctly taking real and imaginary parts of their expanded \((c + \mathrm{i}s)^5\) and dividing correctly (numerator and denominator must contain the correct number of terms and no i’s).
Dependent on both previous M marks
A1: AG - www
Must see mathematically each term in both the numerator and the denominator being divided by \(c^5\) oe (e.g. multiplying by \(c^{-5}\)) so \(\dfrac{c^{-5}\left(5c^4s - 10c^2s^3 + s^5\right)}{c^{-5}\left(c^5 - 10c^3s^2 + 5cs^4\right)}\) or \(\dfrac{\frac{5c^4s - 10c^2s^3 + s^5}{c^5}}{\frac{c^5 - 10c^3s^2 + 5cs^4}{c^5}}\) is fine but \(\dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} \div c^5\) or \(\dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} \div \dfrac{c^5}{c^5}\) etc. or in words (e.g. ‘divide both by \(c^5\)’) is A0
Note that \(\dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} = \dfrac{\frac{5s}{c} - \frac{10s^3}{c^3} + \frac{s^5}{c^5}}{1 - \frac{10s^2}{c^2} + \frac{5s^4}{c^4}}\) is A0 (as AG)
Final answer must be in terms of tan not \(t\)
| Scheme | Marks | AO |
|---|---|---|
| (i) DR \(5t^4 - 10t^2 + 1 = t^5 - 10t^3 + 5t\) \(\Rightarrow t^5 - 5t^4 - 10t^3 + 10t^2 + 5t - 1 \; (= 0)\) | M1* | 2.1 |
| \(\theta = \dfrac{1}{20}\pi, \dfrac{5}{20}\pi, \dfrac{9}{20}\pi, \dfrac{13}{20}\pi, \dfrac{17}{20}\pi\) | B1 | 1.1 |
| \(\tan\left(\dfrac{5}{20}\pi\right) = 1\) so \((t - 1)\) is a factor of \(t^5 - 5t^4 - 10t^3 + 10t^2 + 5t - 1\) \(\Rightarrow (t - 1)\left(t^4 - 4t^3 - 14t^2 - 4t + 1\right)\) | B1dep* | 3.1a |
| \(\tan\left(\dfrac{1}{20}\pi\right), \tan\left(\dfrac{9}{20}\pi\right), \tan\left(\dfrac{13}{20}\pi\right), \tan\left(\dfrac{17}{20}\pi\right)\) | A1 | 2.2a |
| [4] | ||
| (ii) DR \((t - 1)^4 = t^4 - 4t^3 + 6t^2 - 4t + 1\) so if \(t^4 - 4t^3 - 14t^2 - 4t + 1 = 0 \Rightarrow (t - 1)^4 = 20t^2\) | B1 | 2.2a |
| \((t - 1)^2 = \pm\sqrt{20}t\) \(\Rightarrow t^2 - \left(2 \pm \sqrt{20}\right)t + 1 \; (= 0)\) | M1 | 3.1a |
| \(\left(t - \left(1 \pm \sqrt{5}\right)\right)^2 - \left(1 \pm \sqrt{5}\right)^2 + 1 = 0\) \(\Rightarrow t = 1 + \sqrt{5} \pm \sqrt{5 + 2\sqrt{5}}\) or \(t = 1 - \sqrt{5} \pm \sqrt{5 - 2\sqrt{5}}\) \(\tan\left(\dfrac{9}{20}\pi\right)\) is positive and the largest root (of the quartic equation) so \(\tan\left(\dfrac{9}{20}\pi\right) = 1 + \sqrt{5} + \sqrt{5 + 2\sqrt{5}}\) | A1 | 2.3 |
| [3] |
Notes
(b)(i)
M1*: Equates result from part (a) to 1, multiplies through by denominator and then rearranges to get all terms on one side. Condone using tan rather than \(t\) - Allow sign errors only.
B1: Finds all five solutions to \(\tan 5\theta = 1\) in the interval \(0 \lt \theta \lt \pi\) - ignore any solutions outside of this interval but if any incorrect in this interval, then B0
B1dep*: Correct justification that \((t - 1)\) is a factor of the correct quintic in \(t\) using the result that \(\tan\left(\frac{5}{20}\pi\right) = 1\) and re-writing quintic as \((t - 1)\left(t^4 - 4t^3 - 14t^2 - 4t + 1\right)\) oe (e.g. by long division) or re-writes quintic as \((t - 1)\left(t^4 - 4t^3 - 14t^2 - 4t + 1\right)\) oe (e.g. by long division) and relates the factor \((t - 1)\) to \(\tan\left(\frac{5}{20}\pi\right)\) (in essence this mark is for justifying that the root \(\tan\left(\frac{5}{20}\pi\right)\) of the correct quintic equation is not a root of the given quartic equation therefore this mark is dependent on having derived the correct expression \(t^5 - 5t^4 - 10t^3 + 10t^2 + 5t - 1 \; (= 0)\))
A1: These exact four roots only (so must not include \(\tan\left(\frac{5}{20}\pi\right)\)) – this mark is not dependent on the previous B1 mark – a correct quintic equation/expression followed by these four roots only scores M1 A1 only. These four roots only without the correct quintic equation/expression seen is no marks
(b)(ii)
B1: www for correctly finding the value of \(k\) – allow for either \((t - 1)^4 = 20t^2\) or \(k = 20\) stated with no working
M1: For correct three term quadratic expressions(s)/equation(s) (in \(t\)), follow through their positive value of \(k\). Condone missing \(\pm\) - so for either \(t^2 - \left(2 \pm \sqrt{k}\right)t + 1\) or \(t^2 - \left(2 + \sqrt{k}\right)t + 1\) or \(t^2 - \left(2 - \sqrt{k}\right)t + 1\) with their \(k\)
A1: AG
Obtains all four roots www and explains that \(\tan\left(\frac{9}{20}\pi\right)\) is positive and the largest root and so \(\tan\left(\frac{9}{20}\pi\right) = 1 + \sqrt{5} + \sqrt{5 + 2\sqrt{5}}\)
(corrected from the printed mark scheme: the second pair of roots is printed as \(t = 1 - \sqrt{5} \pm \sqrt{5 + 2\sqrt{5}}\); since \(\left(1 - \sqrt{5}\right)^2 - 1 = 5 - 2\sqrt{5}\), it should be \(t = 1 - \sqrt{5} \pm \sqrt{5 - 2\sqrt{5}}\))