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Date May 2009 Marks available 22 Reference code 09M.1.hl.TZ2.12
Level HL only Paper 1 Time zone TZ2
Command term Explain, Find, Show that, State, and Hence Question number 12 Adapted from N/A

Question

The complex number z is defined as z=cosθ+isinθ .

(a)     State de Moivre’s theorem.

(b)     Show that zn1zn=2isin(nθ) .

(c)     Use the binomial theorem to expand (z1z)5 giving your answer in simplified form.

(d)     Hence show that 16sin5θ=sin5θ5sin3θ+10sinθ .

(e)     Check that your result in part (d) is true for θ=π4 .

(f)     Find π20sin5θdθ .

(g)     Hence, with reference to graphs of circular functions, find π20cos5θdθ , explaining your reasoning.

Markscheme

(a)     any appropriate form, e.g. (cosθ+isinθ)n=cos(nθ)+isin(nθ)     A1

[1 mark]

 

(b)     zn=cosnθ+isinnθ     A1

1zn=cos(nθ)+isin(nθ)     (M1)

=cosnθisin(nθ)     A1

therefore zn1zn=2isin(nθ)     AG

[3 marks]

 

(c)     (z1z)5=z5+(51)z4(1z)+(52)z3(1z)2+(53)z2(1z)3+(54)z(1z)4+(1z)5     (M1)(A1)

=z55z3+10z10z+5z31z5     A1

[3 marks]

 

(d)     (z1z)5=z51z55(z31z3)+10(z1z)     M1A1

(2isinθ)5=2isin5θ10isin3θ+20isinθ     M1A1

16sin5θ=sin5θ5sin3θ+10sinθ     AG

[4 marks]

 

(e)     16sin5θ=sin5θ5sin3θ+10sinθ

LHS=16(sinπ4)5

=16(22)5

=22(=42)     A1

RHS=sin(5π4)5sin(3π4)+10sin(π4)

=225(22)+10(22)     M1A1

Note: Award M1 for attempted substitution.

 

=22(=42)     A1

hence this is true for θ=π4     AG

[4 marks]

 

(f)     π20sin5θdθ=116π20(sin5θ5sin3θ+10sinθ)dθ     M1

=116[cos5θ5+5cos3θ310cosθ]π20     A1

=116[0(15+5310)]     A1

=815     A1

[4 marks]

 

(g)     π20cos5θdθ=815 , with appropriate reference to symmetry and graphs.     A1R1R1

Note: Award first R1 for partially correct reasoning e.g. sketches of graphs of sin and cos.

Award second R1 for fully correct reasoning involving sin5 and cos5.

 

[3 marks]

Total [22 marks]

Examiners report

Many students in b) substituted for the second term (again not making the connection to part a)) on the LHS and multiplied by the conjugate, which some managed well but it is inefficient. The binomial expansion was done well even if students did not do the earlier part. The connection between d) and f) was missed by many which lead to some creative attempts at the integral. Very few attempted the last part and of those many attempted another integral, ignoring the hence, while others related to the graph of sin and cos but not to the particular graphs here.

Syllabus sections

Topic 1 - Core: Algebra » 1.7 » Powers of complex numbers: de Moivre’s theorem.
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