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Date May 2017 Marks available 3 Reference code 17M.1.AHL.TZ2.H_11
Level Additional Higher Level Paper Paper 1 Time zone Time zone 2
Command term Show that Question number H_11 Adapted from N/A

Question

Let z = 1 cos 2 θ i sin 2 θ ,   z C ,   0 θ π .

Solve 2 sin ( x + 60 ) = cos ( x + 30 ) ,   0 x 180 .

[5]
a.

Show that sin 105 + cos 105 = 1 2 .

[3]
b.

Find the modulus and argument of z in terms of θ . Express each answer in its simplest form.

[9]
c.i.

Hence find the cube roots of z  in modulus-argument form.

[5]
c.ii.

Markscheme

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

2 sin ( x + 60 ) = cos ( x + 30 )

2 ( sin x cos 60 + cos x sin 60 ) = cos x cos 30 sin x sin 30      (M1)(A1)

2 sin x × 1 2 + 2 cos x × 3 2 = cos x × 3 2 sin x × 1 2      A1

3 2 sin x = 3 2 cos x

tan x = 1 3      M1

x = 150      A1

[5 marks]

a.

EITHER

choosing two appropriate angles, for example 60° and 45°     M1

sin 105 = sin 60 cos 45 + cos 60 sin 45 and

cos 105 = cos 60 cos 45 sin 60 sin 45      (A1)

sin 105 + cos 105 = 3 2 × 1 2 + 1 2 × 1 2 + 1 2 × 1 2 3 2 × 1 2      A1

= 1 2      AG

OR

attempt to square the expression     M1

( sin 105 + cos 105 ) 2 = sin 2 105 + 2 sin 105 cos 105 + cos 2 105

( sin 105 + cos 105 ) 2 = 1 + sin 210      A1

= 1 2      A1

sin 105 + cos 105 = 1 2   AG

 

[3 marks]

b.

EITHER

z = ( 1 cos 2 θ ) i sin 2 θ

| z | = ( 1 cos 2 θ ) 2 + ( sin 2 θ ) 2      M1

| z | = 1 2 cos 2 θ + cos 2 2 θ + sin 2 2 θ      A1

= 2 ( 1 cos 2 θ )      A1

= 2 ( 2 sin 2 θ )

= 2 sin θ      A1

let arg ( z ) = α

tan α = sin 2 θ 1 cos 2 θ      M1

= 2 sin θ cos θ 2 sin 2 θ      (A1)

= cot θ      A1

arg ( z ) = α = arctan ( tan ( π 2 θ ) )      A1

= θ π 2      A1

OR

z = ( 1 cos 2 θ ) i sin 2 θ

= 2 sin 2 θ 2 i sin θ cos θ      M1A1

= 2 sin θ ( sin θ i cos θ )      (A1)

= 2 i sin θ ( cos θ + i sin θ )      M1A1

= 2 sin θ ( cos ( θ π 2 ) + i sin ( θ π 2 ) )      M1A1

| z | = 2 sin θ      A1

arg ( z ) = θ π 2      A1

[9 marks]

c.i.

attempt to apply De Moivre’s theorem     M1

( 1 cos 2 θ i sin 2 θ ) 1 3 = 2 1 3 ( sin θ ) 1 3 [ cos ( θ π 2 + 2 n π 3 ) + i sin ( θ π 2 + 2 n π 3 ) ]      A1A1A1

 

Note:     A1 for modulus, A1 for dividing argument of z by 3 and A1 for 2 n π .

 

Hence cube roots are the above expression when n = 1 ,   0 ,   1 . Equivalent forms are acceptable.     A1

[5 marks]

c.ii.

Examiners report

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a.
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b.
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c.i.
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c.ii.

Syllabus sections

Topic 1—Number and algebra » AHL 1.13—Polar and Euler form
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Topic 1—Number and algebra

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