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 (cos θ+i sin θ)4. Give your answer in the form a+bi where a and b are expressed in terms of sin θ and cos θ.
Use de Moivre’s theorem and the result from part (a) to show that cot 4θ=cot4 θ-6 cot2 θ+14 cot3 θ-4 cot θ.
Use the identity from part (b) to show that the quadratic equation x2-6x+1=0 has roots cot2π8 and cot23π8.
Hence find the exact value of cot23π8.
Deduce a quadratic equation with integer coefficients, having roots cosec2 π8 and cosec2 3π8.
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 (cos θ+i sin θ)4 M1
=C04 cos4 θ+C14 cos3 θ(i sin θ)+C24 cos2 θ(i2 sin2 θ)+C34 cos θ(i3 sin3 θ)+C44(i4 sin4 θ) A1
= (cos4 θ-6 cos2 θ sin2 θ+sin4 θ)+i(4 cos3 θ sin θ-4 cos θ sin3 θ) A1
[3 marks]
(using de Moivre’s theorem with n=4 gives) cos 4θ+i sin 4θ (A1)
equates both the real and imaginary parts of cos 4θ+i sin 4θ and (cos4 θ -6 cos2 θ sin2 θ+sin4 θ)+i(4 cos3 θ sin θ-4 cos θ sin3 θ) M1
cos 4θ=cos4 θ-6 cos2 θ sin2 θ+sin4 θ and sin 4θ=4 cos3 θ sin θ-4 cos θ sin3 θ
recognizes that cot 4θ=cos 4θsin 4θ (A1)
substitutes for sin 4θ and cos 4θ into cos 4θsin 4θ M1
cot 4θ=cos4 θ -6 cos2 θ sin2 θ+sin4 θ4 cos3 θ sin θ-4 cos θ sin3 θ
divides the numerator and denominator by sin4 θ to obtain
cot 4θ=cos4 θ -6 cos2 θ sin2 θ+sin4 θsin4 θ4 cos3 θ sin θ-4 cos θ sin3 θsin4 θ A1
cot 4θ=cot4 θ-6 cot2 θ+14 cot3 θ-4 cot θ AG
[5 marks]
setting cot 4θ=0 and putting x=cot2 θ in the numerator of cot 4θ=cot4 θ-6 cot2 θ+14 cot3 θ-4 cot θ gives x2-6x+1=0 M1
attempts to solve cot 4θ=0 for θ M1
4θ=π2, 3π2, … (A1)
A1
Note: Do not award the final A1 if solutions other than are listed.
finding the roots of corresponds to finding the roots of where R1
so the equation as roots and AG
[5 marks]
attempts to solve for M1
A1
since has the smaller value of the two roots R1
Note: Award R1 for an alternative convincing valid reason.
so A1
[4 marks]
let
uses where (M1)
M1
A1
[3 marks]