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Date November 2014 Marks available 10 Reference code 14N.1.hl.TZ0.13
Level HL only Paper 1 Time zone TZ0
Command term Hence, Show that, State, and Verify Question number 13 Adapted from N/A

Question

(i)     Show that (1+itanθ)n+(1itanθ)n=2cosnθcosnθ,cosθ0.

(ii)     Hence verify that itan3π8 is a root of the equation (1+z)4+(1z)4=0,zC.

(iii)     State another root of the equation (1+z)4+(1z)4=0,zC.

[10]
a.

(i)     Use the double angle identity tan2θ=2tanθ1tan2θ to show that tanπ8=21.

(ii)     Show that cos4x=8cos4x8cos2x+1.

(iii)     Hence find the value of π802cos4xcos2xdx.

[13]
b.

Markscheme

(i)     METHOD 1

(1+itanθ)n+(1itanθ)n=(1+isinθcosθ)n+(1isinθcosθ)n     M1

=(cosθ+isinθcosθ)n+(cosθisinθcosθ)n     A1

by de Moivre’s theorem     (M1)

(cosθ+isinθcosθ)n=cosnθ+isinnθcosnθ     A1

recognition that cosθisinθ is the complex conjugate of cosθ+isinθ     (R1)

use of the fact that the operation of complex conjugation commutes with the operation of raising to an integer power:

(cosθisinθcosθ)n=cosnθisinnθcosnθ     A1

(1+itanθ)n+(1itanθ)n=2cosnθcosnθ     AG

METHOD 2

(1+itanθ)n+(1itanθ)n=(1+itanθ)n+(1+itan(θ))n     (M1)

=(cosθ+isinθ)ncosnθ+(cos(θ)+isin(θ))ncosnθ     M1A1

 

Note:     Award M1 for converting to cosine and sine terms.

 

use of de Moivre’s theorem     (M1)

=1cosnθ(cosnθ+isinnθ+cos(nθ)+isin(nθ))     A1

=2cosnθcos2θascos(nθ)=cosnθandsin(nθ)=sinnθ     R1AG

(ii)     (1+itan3π8)4+(1itan3π8)4=2cos(4×3π8)cos43π8     (A1)

=2cos3π2cos43π8     A1

=0ascos3π2=0     R1

 

Note:     The above working could involve theta and the solution of cos(4θ)=0.

 

so itan3π8 is a root of the equation     AG

(iii)     either itan3π8oritanπ8oritanπ8     A1

 

Note:     Accept itan5π8oritan7π8.

Accept (1+2)ior(12)ior(1+2)i.

[10 marks]

a.

(i)     tanπ4=2tanπ81tan2π8     (M1)

tan2π8+2tanπ81=0     A1

let t=tanπ8

attempting to solve t2+2t1=0fort     M1

t=1±2     A1

π8 is a first quadrant angle and tan is positive in this quadrant, so

tanπ8>0     R1

tanπ8=21     AG

(ii)     cos4x=2cos22x1     A1

=2(2cos2x1)21     M1

=2(4cos4x4cos2x+1)1     A1

=8cos4x8cos2x+1     AG

 

Note:     Accept equivalent complex number derivation.

 

(iii)     π802cos4xcos2xdx=2π808cos4x8cos2x+1cos2xdx

=2π808cos2x8+sec2xdx     M1

 

Note:     The M1 is for an integrand involving no fractions.

 

use of cos2x=12(cos2x+1)     M1

=2π804cos2x4+sec2xdx     A1

=[4sin2x8x+2tanx]π80     A1

=42π2(or equivalent)     A1

[13 marks]

Total [23 marks]

b.

Examiners report

Fairly successful.

a.

(i)     Most candidates attempted to use the hint. Those who doubled the angle were usually successful – but many lost the final mark by not giving a convincing reason to reject the negative solution to the intermediate quadratic equation. Those who halved the angle got nowhere.

(ii)     The majority of candidates obtained full marks.

(iii)     This was poorly answered, few candidates realising that part of the integrand could be re-expressed using 1cos2x=sec2x, which can be immediately integrated.

b.

Syllabus sections

Topic 1 - Core: Algebra » 1.7 » Powers of complex numbers: de Moivre’s theorem.
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