Date | November 2018 | Marks available | 5 | Reference code | 18N.1.AHL.TZ0.H_11 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Express and Find | Question number | H_11 | Adapted from | N/A |
Question
Let S be the sum of the roots found in part (a).
Find the roots of z24=1 which satisfy the condition 0<arg(z)<π2, expressing your answers in the form reiθ, where r, θ∈R+.
Show that Re S = Im S.
By writing π12 as (π4−π6), find the value of cos π12 in the form √a+√bc, where a, b and c are integers to be determined.
Hence, or otherwise, show that S = 12(1+√2)(1+√3)(1+i).
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(r(cosθ+isinθ))24=1(cos0+isin0)
use of De Moivre’s theorem (M1)
r24=1⇒r=1 (A1)
24θ=2πn⇒θ=πn12, (n∈Z) (A1)
0<arg(z)<π2⇒n= 1, 2, 3, 4, 5
z=eπi12 or e2πi12 or e3πi12 or e4πi12 or e5πi12 A2
Note: Award A1 if additional roots are given or if three correct roots are given with no incorrect (or additional) roots.
[5 marks]
Re S = cosπ12+cos2π12+cos3π12+cos4π12+cos5π12
Im S = sinπ12+sin2π12+sin3π12+sin4π12+sin5π12 A1
Note: Award A1 for both parts correct.
but sin5π12=cosπ12, sin4π12=cos2π12, sin3π12=cos3π12, sin2π12=cos4π12 and sinπ12=cos5π12 M1A1
⇒ Re S = Im S AG
Note: Accept a geometrical method.
[4 marks]
cosπ12=cos(π4−π6)=cosπ4cosπ6+sinπ4sinπ6 M1A1
=√22√32+√2212
=√6+√24 A1
[3 marks]
cos5π12=cos(π6+π4)=cosπ6cosπ4−sinπ6sinπ4 (M1)
Note: Allow alternative methods eg cos5π12=sinπ12=sin(π4−π6).
=√32√22−12√22=√6−√24 (A1)
Re S = cosπ12+cos2π12+cos3π12+cos4π12+cos5π12
Re S = √2+√64+√32+√22+12+√6−√24 A1
=12(√6+1+√2+√3) A1
=12(1+√2)(1+√3)
S = Re(S)(1 + i) since Re S = Im S, R1
S = 12(1+√2)(1+√3)(1+i) AG
[4 marks]