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 zn−1zn=2isin(nθ) .
(c) Use the binomial theorem to expand (z−1z)5 giving your answer in simplified form.
(d) Hence show that 16sin5θ=sin5θ−5sin3θ+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)
=cosnθ−isin(nθ) A1
therefore zn−1zn=2isin(nθ) AG
[3 marks]
(c) (z−1z)5=z5+(51)z4(−1z)+(52)z3(−1z)2+(53)z2(−1z)3+(54)z(−1z)4+(−1z)5 (M1)(A1)
=z5−5z3+10z−10z+5z3−1z5 A1
[3 marks]
(d) (z−1z)5=z5−1z5−5(z3−1z3)+10(z−1z) M1A1
(2isinθ)5=2isin5θ−10isin3θ+20isinθ M1A1
16sin5θ=sin5θ−5sin3θ+10sinθ AG
[4 marks]
(e) 16sin5θ=sin5θ−5sin3θ+10sinθ
LHS=16(sinπ4)5
=16(√22)5
=2√2(=4√2) A1
RHS=sin(5π4)−5sin(3π4)+10sin(π4)
=−√22−5(√22)+10(√22) M1A1
Note: Award M1 for attempted substitution.
=2√2(=4√2) A1
hence this is true for θ=π4 AG
[4 marks]
(f) ∫π20sin5θdθ=116∫π20(sin5θ−5sin3θ+10sinθ)dθ M1
=116[−cos5θ5+5cos3θ3−10cosθ]π20 A1
=116[0−(−15+53−10)] 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.