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)cosx−cos((2n+1)x)sinxsin2nx=sin((2n+1)x)cosx−cos((2n+1)x)sinx.
(b) Hence prove, by induction, that
cosx+cos3x+cos5x+…+cos((2n−1)x)=sin2nx2sinx,cosx+cos3x+cos5x+…+cos((2n−1)x)=sin2nx2sinx,
for all n∈Z+, sinx≠0.
(c) Solve the equation cosx+cos3x=12, 0<x<π.
Markscheme
(a) sin(2n+1)xcosx−cos(2n+1)xsinx=sin(2n+1)x−x 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(2k−1)x=sin2kx2sinx
if n = k + 1 then
cosx+cos3x+cos5x+…+cos(2k−1)x+cos(2k+1)x M1
=sin2kx2sinx+cos(2k+1)x A1
=sin2kx+2cos(2k+1)xsinx2sinx M1
=sin(2k+1)xcosx−cos(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 n∈Z+ R1
Note: Final R1 is independent of previous work.
[12 marks]
(c) sin4x2sinx=12 M1A1
sin4x=sinx
4x=x⇒x=0 but this is impossible
4x=π−x⇒x=π5 A1
4x=2π+x⇒x=2π3 A1
4x=3π−x⇒x=3π5 A1
for not including any answers outside the domain R1
Note: Award the first M1A1 for correctly obtaining 8cos3x−4cosx−1=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.