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Date November Example questions Marks available 3 Reference code EXN.1.AHL.TZ0.12
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Deduce Question number 12 Adapted from N/A

Question

Use the binomial theorem to expand cosθ+isinθ4. Give your answer in the form a+bi where a and b are expressed in terms of sinθ and cosθ.

[3]
a.

Use de Moivre’s theorem and the result from part (a) to show that cot4θ=cot4θ-6cot2θ+14cot3θ-4cotθ.

[5]
b.

Use the identity from part (b) to show that the quadratic equation x2-6x+1=0 has roots cot2π8 and cot23π8.

[5]
c.

Hence find the exact value of cot23π8.

[4]
d.

Deduce a quadratic equation with integer coefficients, having roots cosec2π8 and cosec23π8.

[3]
e.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

uses the binomial theorem on cosθ+isinθ4       M1

=C04cos4θ+C14cos3θisinθ+C24cos2θi2sin2θ+C34cosθi3sin3θ+C44i4sin4θ        A1

=cos4θ-6cos2θsin2θ+sin4θ+i4cos3θsinθ-4cosθsin3θ        A1

 

[3 marks]

a.

(using de Moivre’s theorem with n=4 gives) cos4θ+isin4θ        (A1)

equates both the real and imaginary parts of cos4θ+isin4θ and cos4θ-6cos2θsin2θ+sin4θ+i4cos3θsinθ-4cosθsin3θ       M1

cos4θ=cos4θ-6cos2θsin2θ+sin4θ  and  sin4θ=4cos3θsinθ-4cosθsin3θ

recognizes that cot4θ=cos4θsin4θ        (A1)

substitutes for sin4θ and cos4θ into cos4θsin4θ       M1

cot4θ=cos4θ-6cos2θsin2θ+sin4θ4cos3θsinθ-4cosθsin3θ

divides the numerator and denominator by sin4θ to obtain

cot4θ=cos4θ-6cos2θsin2θ+sin4θsin4θ4cos3θsinθ-4cosθsin3θsin4θ        A1

cot4θ=cot4θ-6cot2θ+14cot3θ-4cotθ        AG

 

[5 marks]

b.

setting cot4θ=0 and putting x=cot2θ in the numerator of cot4θ=cot4θ-6cot2θ+14cot3θ-4cotθ gives x2-6x+1=0        M1

attempts to solve cot4θ=0 for θ        M1

4θ=π2,3π2, 4θ=122n+1π,n=0,1,        (A1)

θ=π8,3π8        A1

 

Note: Do not award the final A1 if solutions other than θ=π8,3π8 are listed.

 

finding the roots of cot4θ=0 θ=π8,3π8 corresponds to finding the roots of x2-6x+1=0 where x=cot2θ        R1

so the equation x2-6x+1=0 as roots cot2π8 and cot23π8        AG

 

[5 marks]

c.

attempts to solve x2-6x+1=0 for x        M1

x=3±22        A1

since cot2π8>cot23π8, cot23π8 has the smaller value of the two roots        R1

 

Note: Award R1 for an alternative convincing valid reason.

 

so cot23π8=3-22        A1

 

[4 marks]

d.

let y=cosec2θ

uses cot2θ=cosec2θ-1 where x=cot2θ        (M1)

x2-6x+1=0y-12-6y-1+1=0        M1

y2-8y+8=0        A1

 

[3 marks]

e.

Examiners report

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Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.9—Reciprocal trig ratios and their pythagorean identities. Inverse circular functions
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Topic 3— Geometry and trigonometry

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