| Date | November Example questions | Marks available | 8 | Reference code | EXN.1.AHL.TZ0.9 |
| Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
| Command term | Prove | Question number | 9 | Adapted from | N/A |
Question
It is given that 2 cos A sin B≡sin(A+B)-sin(A-B). (Do not prove this identity.)
Using mathematical induction and the above identity, prove that nΣr=1cos(2r-1)θ=sin 2nθ2 sin θ for n∈ℤ+.
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.
let P(n) be the proposition that nΣr=1cos(2r-1)θ=sin 2nθ2 sin θ for n∈ℤ+
considering P(1):
LHS=cos(1)θ=cos θ and RHS=sin 2θ2 sin θ=2 sin θ cosθ2 sin θ=cosθ=LHS
so P(1) is true R1
assume P(k) is true, i.e. kΣr=1cos(2r-1)θ=sin 2kθ2 sin θ (k∈ℤ+) M1
Note: Award M0 for statements such as “let n=k”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.
considering P(k+1)
k+1Σr=1cos(2r-1)θ=kΣr=1cos(2r-1)θ+cos(2(k+1)-1)θ M1
=sin 2kθ2 sin θ +cos(2(k+1)-1)θ A1
=sin 2kθ+2 cos((2k+1)θ) sin θ2 sin θ
=sin 2kθ+sin((2k+1)θ+θ)- sin((2k+1)θ-θ)2 sin θ M1
Note: Award M1 for use of 2 cos A sin B=sin(A+B)-sin(A-B) with A=(2k+1)θ and B=θ.
=sin 2kθ+sin(2k+2)θ- sin 2kθ2 sin θ A1
=sin 2(k+1)θ2 sin θ A1
P(k+1) is true whenever P(k) is true, P(1) is true, so P(n) is true for n∈ℤ+ R1
Note: Award the final R1 mark provided at least five of the previous marks have been awarded.
[8 marks]
