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Date May 2010 Marks available 20 Reference code 10M.1.hl.TZ1.13
Level HL only Paper 1 Time zone TZ1
Command term Hence, Prove, Show that, and Solve Question number 13 Adapted from N/A

Question

(a)     Show that sin2nx=sin((2n+1)x)cosxcos((2n+1)x)sinxsin2nx=sin((2n+1)x)cosxcos((2n+1)x)sinx.

(b)     Hence prove, by induction, that

cosx+cos3x+cos5x++cos((2n1)x)=sin2nx2sinx,cosx+cos3x+cos5x++cos((2n1)x)=sin2nx2sinx,

for all nZ+sinx0.

(c)     Solve the equation cosx+cos3x=12, 0<x<π.

Markscheme

(a)     sin(2n+1)xcosxcos(2n+1)xsinx=sin(2n+1)xx     M1A1

=sin2nx     AG

[2 marks]

 

(b)     if n = 1     M1

LHS=cosx

RHS=sin2x2sinx=2sinxcosx2sinx=cosx     M1

so LHS = RHS and the statement is true for n = 1     R1

assume true for n = k     M1

Note: Only award M1 if the word true appears.

   Do not award M1 for ‘let n = k’ only.

   Subsequent marks are independent of this M1.

 

so cosx+cos3x+cos5x++cos(2k1)x=sin2kx2sinx

if n = k + 1 then

cosx+cos3x+cos5x++cos(2k1)x+cos(2k+1)x     M1

=sin2kx2sinx+cos(2k+1)x     A1

=sin2kx+2cos(2k+1)xsinx2sinx     M1

=sin(2k+1)xcosxcos(2k+1)xsinx+2cos(2k+1)xsinx2sinx     M1

=sin(2k+1)xcosx+cos(2k+1)xsinx2sinx     A1

=sin(2k+2)x2sinx     M1

=sin2(k+1)x2sinx     A1

so if true for n = k, then also true for n = k + 1

as true for n = 1 then true for all nZ+     R1

Note: Final R1 is independent of previous work.

[12 marks]

 

(c)     sin4x2sinx=12     M1A1

sin4x=sinx

4x=xx=0 but this is impossible

4x=πxx=π5     A1

4x=2π+xx=2π3     A1

4x=3πxx=3π5     A1

for not including any answers outside the domain     R1

Note: Award the first M1A1 for correctly obtaining 8cos3x4cosx1=0 or equivalent and subsequent marks as appropriate including the answers (12,1±54).

[6 marks]

 

Total [20 marks]

Examiners report

This question showed the weaknesses of many candidates in dealing with formal proofs and showing their reasoning in a logical manner. In part (a) just a few candidates clearly showed the result and part (b) showed that most candidates struggle with the formality of a proof by induction. The logic of many solutions was poor, though sometimes contained correct trigonometric work. Very few candidates were successful in answering part (c) using the unit circle. Most candidates attempted to manipulate the equation to obtain a cubic equation but made little progress. A few candidates guessed 2π3 as a solution but were not able to determine the other solutions.

Syllabus sections

Topic 1 - Core: Algebra » 1.4 » Proof by mathematical induction.

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