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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 2cosAsinBsinA+B-sinA-B. (Do not prove this identity.)

Using mathematical induction and the above identity, prove that Σr=1ncos2r-1θ=sin2nθ2sinθ for n+.

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 Pn be the proposition that Σr=1ncos2r-1θ=sin2nθ2sinθ for n+

considering P1:

LHS=cos1θ=cosθ and RHS=sin2θ2sinθ=2sinθcosθ2sinθ=cosθ=LHS

so P1 is true        R1

assume Pk is true, i.e. Σr=1kcos2r-1θ=sin2kθ2sinθ k+        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 Pk+1

Σr=1k+1cos2r-1θ=Σr=1kcos2r-1θ+cos2k+1-1θ        M1

=sin2kθ2sinθ +cos2k+1-1θ        A1

=sin2kθ+2cos2k+1θsinθ2sinθ 

=sin2kθ+sin2k+1θ+θ-sin2k+1θ-θ2sinθ         M1

 

Note: Award M1 for use of 2cosAsinB=sinA+B-sinA-B with A=2k+1θ and B=θ.

 

=sin2kθ+sin2k+2θ-sin2kθ2sinθ         A1

=sin2k+1θ2sinθ         A1

Pk+1 is true whenever Pk is true, P1 is true, so Pn is true for n+        R1

 

Note: Award the final R1 mark provided at least five of the previous marks have been awarded.

 

[8 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Proof by induction, contradiction, counterexamples
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Topic 1—Number and algebra

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