FP2 June 2014 (R) Q7

EdexcelOld spec14 marksDe Moivre's Theorem

(a) Use de Moivre’s theorem to show that \[\sin 5\theta \equiv 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\] (5)

(b) Hence find the five distinct solutions of the equation \[16x^5 - 20x^3 + 5x + \frac{1}{2} = 0\] giving your answers to 3 decimal places where necessary. (5)

(c) Use the identity given in (a) to find \[\int_0^{\frac{\pi}{4}} \left(4\sin^5\theta - 5\sin^3\theta\right)\mathrm{d}\theta\] expressing your answer in the form \(a\sqrt{2} + b\), where \(a\) and \(b\) are rational numbers. (4)

Mark scheme (a)

Scheme	Marks
\(\left(\cos\theta + \mathrm{i}\sin\theta\right)^5 = \cos 5\theta + \mathrm{i}\sin 5\theta\)	B1
\(= \cos^5\theta + 5\cos^4\theta\left(\mathrm{i}\sin\theta\right) + \dfrac{5\times4}{2!}\cos^3\theta\left(\mathrm{i}\sin\theta\right)^2\) \(+ \dfrac{5\times4\times3}{3!}\cos^2\theta\left(\mathrm{i}\sin\theta\right)^3 + \dfrac{5\times4\times3\times2}{4!}\cos\theta\left(\mathrm{i}\sin\theta\right)^4 + \left(\mathrm{i}\sin\theta\right)^5\)	M1
\(= \cos^5\theta + 5\mathrm{i}\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta\) \(-10\mathrm{i}\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + \mathrm{i}\sin^5\theta\)	A1
\(\sin 5\theta = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\)
\(= 5\left(1 - \sin^2\theta\right)^2\sin\theta - 10\left(1 - \sin^2\theta\right)\sin^3\theta + \sin^5\theta\)	M1
\(\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta\) *	A1
	(5)

Notes

B1 applies de Moivre correctly
M1 uses binomial theorem to expand \(\left(\cos\theta + \mathrm{i}\sin\theta\right)^5\) May only show imaginary parts - ignore errors in real part
A1 simplifies coefficients to obtain a simplified result with all imaginary terms correct
M1 equates imaginary parts and obtains an expression for \(\sin 5\theta\) in terms of powers of \(\sin\theta\)
A1 cso correct result

(Corrected from the printed mark scheme: the second term of the expansion is printed as \(5\cos^4\left(\mathrm{i}\sin\theta\right)\), without the \(\theta\).)

Mark scheme (b)

Scheme	Marks
Let \(x = \sin\theta \quad 16x^5 - 20x^3 + 5x = -\dfrac{1}{2} \Rightarrow \sin 5\theta = -\dfrac{1}{2}\)	M1
\(5\theta = 210,\ 330,\ 570,\ 690,\ 930,\ 1050,\ 1290\) (or in radians) Or 210, 570, 930, 1290, 1650	A1, A1
\(\theta = 42,\ 66,\ (114),\ (138),\ 186,\ 210,\ 258\) (or in radians) Or 42, 114, 186, 258, 330	dM1(at least 2 values)
\(\sin\theta = 0.669,\ 0.914,\ -0.105,\ -0.5,\ -0.978\)	A1
	(5)

Notes

M1 uses substitution \(x = \sin\theta\) deduces that \(\sin 5\theta = \pm\dfrac{1}{2}\)
A1A1 gives a set of results for \(5\theta\) - A1 for 3 useable results A1 for the remaining 2 useable results (no repeats in the set of 5)
M1 at least 2 values for \(\theta\)
A1 for the 5 different values of \(x\)

Mark scheme (c)

Scheme	Marks
\(\displaystyle\int_0^{\frac{\pi}{4}}\left(4\sin^5\theta - 5\sin^3\theta\right)\mathrm{d}\theta = \frac{1}{4}\int_0^{\frac{\pi}{4}}\left(\sin 5\theta - 5\sin\theta\right)\mathrm{d}\theta\)	M1
\(= \dfrac{1}{4}\left[-\dfrac{1}{5}\cos 5\theta + 5\cos\theta\right]_0^{\frac{\pi}{4}}\)	A1
\(\dfrac{1}{4}\left[-\dfrac{1}{5}\cos\dfrac{5\pi}{4} + 5\cos\dfrac{\pi}{4} - \left(-\dfrac{1}{5} + 5\right)\right]\)
\(= \dfrac{1}{4}\left[\dfrac{1}{5}\times\dfrac{1}{\sqrt{2}} + \dfrac{5}{\sqrt{2}} - 4\dfrac{4}{5}\right]\)	M1
\(= \dfrac{13\sqrt{2}}{20} - \dfrac{6}{5}\)	A1
	(4)
	(14 marks)

Notes

M1 uses previous work to change the integrand
A1 correct result after integrating - limits can be ignored
M1 substitute given limits and use numerical values for trig functions
A1 final answer correct (oe provided in the given form)