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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  z 24 = 1 which satisfy the condition 0 < arg ( z ) < π 2 , expressing your answers in the form r e i θ , where r , θ R + .

[5]
a.

Show that Re S = Im S.

[4]
b.i.

By writing  π 12 as ( π 4 π 6 ) , find the value of cos  π 12 in the form a + b c , where a b and  c are integers to be determined.

[3]
b.ii.

Hence, or otherwise, show that S = 1 2 ( 1 + 2 ) ( 1 + 3 ) ( 1 + i ) .

[4]
b.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

( r ( cos θ + i sin θ ) ) 24 = 1 ( cos 0 + i sin 0 )

use of De Moivre’s theorem       (M1)

r 24 = 1 r = 1       (A1)

24 θ = 2 π n θ = π n 12 ( n Z )       (A1)

0 < arg ( z ) < π 2 n = 1, 2, 3, 4, 5

z = e π i 12 or  e 2 π i 12 or  e 3 π i 12 or  e 4 π i 12 or  e 5 π i 12       A2

Note: Award A1 if additional roots are given or if three correct roots are given with no incorrect (or additional) roots.

 

[5 marks]

a.

Re S =  cos π 12 + cos 2 π 12 + cos 3 π 12 + cos 4 π 12 + cos 5 π 12

Im S =  sin π 12 + sin 2 π 12 + sin 3 π 12 + sin 4 π 12 + sin 5 π 12       A1

Note: Award A1 for both parts correct.

but  sin 5 π 12 = cos π 12 ,   sin 4 π 12 = cos 2 π 12 ,   sin 3 π 12 = cos 3 π 12 ,   sin 2 π 12 = cos 4 π 12 and  sin π 12 = cos 5 π 12       M1A1

⇒ Re S = Im S       AG

Note: Accept a geometrical method.

 

[4 marks]

b.i.

cos π 12 = cos ( π 4 π 6 ) = cos π 4 cos π 6 + sin π 4 sin π 6       M1A1

= 2 2 3 2 + 2 2 1 2

= 6 + 2 4        A1

 

[3 marks]

b.ii.

 

cos 5 π 12 = cos ( π 6 + π 4 ) = cos π 6 cos π 4 sin π 6 sin π 4       (M1)

Note: Allow alternative methods eg  cos 5 π 12 = sin π 12 = sin ( π 4 π 6 ) .

= 3 2 2 2 1 2 2 2 = 6 2 4       (A1)

Re S =  cos π 12 + cos 2 π 12 + cos 3 π 12 + cos 4 π 12 + cos 5 π 12

Re S =  2 + 6 4 + 3 2 + 2 2 + 1 2 + 6 2 4       A1

= 1 2 ( 6 + 1 + 2 + 3 )       A1

= 1 2 ( 1 + 2 ) ( 1 + 3 )

S = Re(S)(1 + i) since Re S = Im S,      R1

S =  1 2 ( 1 + 2 ) ( 1 + 3 ) ( 1 + i )       AG

 

[4 marks]

b.iii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.10—Compound angle identities
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Topic 1—Number and algebra » AHL 1.13—Polar and Euler form
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