IB DP Maths Topic 1.3 :The binomial theorem HL Paper 2

(i)     \({(\cos \theta  + {\text{i}}\sin \theta )^5}\)

\( = {\cos ^5}\theta  + 5{\text{i}}{\cos ^4}\theta \sin \theta  + 10{{\text{i}}^2}{\cos ^3}\theta {\sin ^2}\theta  + \)

\(10{{\text{i}}^3}{\cos ^2}\theta {\sin ^3}\theta  + 5{{\text{i}}^4}\cos \theta {\sin ^4}\theta  + {{\text{i}}^5}{\sin ^5}\theta \)     A1A1

\(( = {\cos ^5}\theta  + 5{\text{i}}{\cos ^4}\theta \sin \theta  – 10{\cos ^3}\theta {\sin ^2}\theta  – \)

\(10{\text{i}}{\cos ^2}\theta {\sin ^3}\theta  + 5\cos \theta {\sin ^4}\theta  + {\text{i}}{\sin ^5}\theta )\)

Note:     Award first A1 for correct binomial coefficients.

(ii)     \({({\text{cis}}\theta )^5} = {\text{cis}}5\theta  = \cos 5\theta  + {\text{i}}\sin 5\theta \)     M1

\( = {\cos ^5}\theta  + 5{\text{i}}{\cos ^4}\theta \sin \theta  – 10{\cos ^3}\theta {\sin ^2}\theta  – 10{\text{i}}{\cos ^2}\theta {\sin ^3}\theta  + \)

\(5\cos \theta {\sin ^4}\theta  + {\text{i}}{\sin ^5}\theta \)     A1

Note:     Previous line may be seen in (i)

equating imaginary terms     M1

\(\sin 5\theta  = 5{\cos ^4}\theta \sin \theta  – 10{\cos ^2}\theta {\sin ^3}\theta  + {\sin ^5}\theta \)     AG

(iii)     equating real terms

\(\cos 5\theta  = {\cos ^5}\theta  – 10{\cos ^3}\theta {\sin ^2}\theta  + 5\cos \theta {\sin ^4}\theta \)     A1

[6 marks]

\({(r{\text{cis}}\alpha )^5} = 1 \Rightarrow {r^5}{\text{cis}}5\alpha  = 1{\text{cis}}0\)     M1

\({r^5} = 1 \Rightarrow r = 1\)     A1

\(5\alpha  = 0 \pm 360k,{\text{ }}k \in \mathbb{Z} \Rightarrow a = 72k\)     (M1)

\(\alpha  = 72^\circ \)     A1

Note:     Award M1A0 if final answer is given in radians.

[4 marks]

use of \(\sin (5 \times 72) = 0\) OR the imaginary part of \(1\) is \(0\)     (M1)

\(0 = 5{\cos ^4}\alpha \sin \alpha  – 10{\cos ^2}\alpha {\sin ^3}\alpha  + {\sin ^5}\alpha \)     A1

\(\sin \alpha  \ne 0 \Rightarrow 0 = 5{(1 – {\sin ^2}\alpha )^2} – 10(1 – {\sin ^2}\alpha ){\sin ^2}\alpha  + {\sin ^4}\alpha \)     M1

Note:     Award M1 for replacing \({\cos ^2}\alpha \).

\(0 = 5(1 – 2{\sin ^2}\alpha  + {\sin ^4}\alpha ) – 10{\sin ^2}\alpha  + 10{\sin ^4}\alpha  + {\sin ^4}\alpha \)     A1

Note:     Award A1 for any correct simplification.

so \(16{\sin ^4}\alpha  – 20{\sin ^2}\alpha  + 5 = 0\)     AG

[4 marks]

\({\sin ^2}\alpha  = \frac{{20 \pm \sqrt {400 – 320} }}{{32}}\)     M1A1

\(\sin \alpha  =  \pm \sqrt {\frac{{20 \pm \sqrt {80} }}{{32}}} \)

\(\sin \alpha  = \frac{{ \pm \sqrt {10 \pm 2\sqrt 5 } }}{4}\)     A1

Note:     Award A1 regardless of signs. Accept equivalent forms with integral denominator, simplification may be seen later.

as \(72 > 60\), \(\sin 72 > \frac{{\sqrt 3 }}{2} = 0.866 \ldots \) we have to take both positive signs (or equivalent argument)     R1

Note:     Allow verification of correct signs with calculator if clearly stated

\(\sin 72 = \frac{{\sqrt {10 + 2\sqrt 5 } }}{4}\)     A1

[5 marks]

Total [19 marks]

In part (i) many candidates tried to multiply it out the binomials rather than using the binomial theorem. In parts (ii) and (iii) many candidates showed poor understanding of complex numbers and made no attempt to equate real and imaginary parts. In a some cases the correct answer to part (iii) was seen although it was unclear how it was obtained.

This question was poorly done. Very few candidates made a good attempt to apply De Moivre’s theorem and most of them could not even equate the moduli to obtain \(r\).

This question was poorly done. From the few candidates that attempted it, many candidates started by writing down what they were trying to prove and made no progress.

Very few made a serious attempt to answer this question. Also very few realised that they could use the answers given in part (c) to attempt this part.