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+(1−itanθ)n=2cosnθcosnθ,cosθ≠0.
(ii) Hence verify that itan3π8 is a root of the equation (1+z)4+(1−z)4=0,z∈C.
(iii) State another root of the equation (1+z)4+(1−z)4=0,z∈C.
(i) Use the double angle identity tan2θ=2tanθ1−tan2θ to show that tanπ8=√2−1.
(ii) Show that cos4x=8cos4x−8cos2x+1.
(iii) Hence find the value of ∫π802cos4xcos2xdx.
Markscheme
(i) METHOD 1
(1+itanθ)n+(1−itanθ)n=(1+isinθcosθ)n+(1−isinθcosθ)n M1
=(cosθ+isinθcosθ)n+(cosθ−isinθcosθ)n A1
by de Moivre’s theorem (M1)
(cosθ+isinθcosθ)n=cosnθ+isinnθcosnθ A1
recognition that cosθ−isinθ 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:
(cosθ−isinθcosθ)n=cosnθ−isinnθcosnθ A1
(1+itanθ)n+(1−itanθ)n=2cosnθcosnθ AG
METHOD 2
(1+itanθ)n+(1−itanθ)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+(1−itan3π8)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 −itan3π8or−itanπ8oritanπ8 A1
Note: Accept itan5π8oritan7π8.
Accept −(1+√2)ior(1−√2)ior(−1+√2)i.
[10 marks]
(i) tanπ4=2tanπ81−tan2π8 (M1)
tan2π8+2tanπ8−1=0 A1
let t=tanπ8
attempting to solve t2+2t−1=0fort M1
t=−1±√2 A1
π8 is a first quadrant angle and tan is positive in this quadrant, so
tanπ8>0 R1
tanπ8=√2−1 AG
(ii) cos4x=2cos22x−1 A1
=2(2cos2x−1)2−1 M1
=2(4cos4x−4cos2x+1)−1 A1
=8cos4x−8cos2x+1 AG
Note: Accept equivalent complex number derivation.
(iii) ∫π802cos4xcos2xdx=2∫π808cos4x−8cos2x+1cos2xdx
=2∫π808cos2x−8+sec2xdx M1
Note: The M1 is for an integrand involving no fractions.
use of cos2x=12(cos2x+1) M1
=2∫π804cos2x−4+sec2xdx A1
=[4sin2x−8x+2tanx]π80 A1
=4√2−π−2(or equivalent) A1
[13 marks]
Total [23 marks]
Examiners report
Fairly successful.
(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.