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HSC 2013 MX2 Marathon (archive) (6 Viewers)

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Sy123

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Re: HSC 2013 4U Marathon

a) Find, in modulus-argument form, the roots of the equation z^(2n+1) =1.
b) Hence factorise z^2n + z^(2n-1) + ... + z^2 + z +1 into quadratic factors with real coefficients.
c) Deduce that 2^n * sin(pi/(2n+1)) * sin(2pi/(2n+1)) * sin(3pi/(2n+1)) ... sin(n*pi/(2n+1)) = sqrt(2n+1)












 
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HeroicPandas

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Re: HSC 2013 4U Marathon

Nice work.


=========








let P(x) = x^2-px +q =0

if a and b are roots, therefore P(a) = P(b)=0

P(a) = a^2 -pa+q=0, times by a^n
P(b) = b^2 -pb +q=0, times by b^n

therefore:
a^(n+2) -pa^(n+1) +qa^n =0
b^(n+2) -pb^(n+1) +qb^n =0

solving simultaneously by ADDITION:
a^(n+2) +b^(n+2) -p(a^(n+1) +b^(n+1)) +q(a^n +b^n) =0

S_n =a^n +b^n
S_n+1 =a^(n+1) +b^(n+1)
S_n+2 =a^(n+2) +b^(n+2)

therefore S_n+2 -pS_n+1 + qS_n =0

ALTERNATIVELY,

since a and b are roots of P(x) = 0, then P(a)=P(b)=0

P(a) = a^2 -pa +q =0............[1]
P(b) = b^2 -pb +q =0.............[2]

S_n+2 -pS_n+1 + qS_n =0
a^(n+2) +b^(n+2) -p[a^(n+1) +b^(n+1)] +q(a^n +b^n) =0

a^(n+2) -pa^(n+1) +qa^n +b^(n+2) -pb^(n+1) +qb^n =0

a^n[a^2 -pa +q] + b^n[b^2 -pb +q] =0

by applying [1] and [2]
0+0=0
LHS=RHS therefore, S_n+2 -pS_n+1 + qS_n =0
 
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Sy123

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Re: HSC 2013 4U Marathon

let P(x) = x^2-px +q =0

if a and b are roots, therefore P(a) = P(b)=0

P(a) = a^2 -pa+q=0, times by a^n
P(b) = b^2 -pb +q=0, times by b^n

therefore:
a^(n+2) -pa^(n+1) +qa^n =0
b^(n+2) -pb^(n+1) +qb^n =0

solving simultaneously by ADDITION:
a^(n+2) +b^(n+2) -p(a^(n+1) +b^(n+1)) +q(a^n +b^n) =0

S_n =a^n +b^n
S_n+1 =a^(n+1) +b^(n+1)
S_n+2 =a^(n+2) +b^(n+2)

therefore S_n+2 -pS_n+1 + qS_n =0

ALTERNATIVELY,

since a and b are roots of P(x) = 0, then P(a)=P(b)=0

P(a) = a^2 -pa +q =0............[1]
P(b) = b^2 -pb +q =0.............[2]

S_n+2 -pS_n+1 + qS_n =0
a^(n+2) +b^(n+2) -p[a^(n+1) +b^(n+1)] +q(a^n +b^n) =0

a^(n+2) -pa^(n+1) +qa^n +b^(n+2) -pb^(n+1) +qb^n =0

a^n[a^2 -pa +q] + b^n[b^2 -pb +q] =0

by applying [1] and [2]
0+0=0
LHS=RHS therefore, S_n+2 -pS_n+1 + qS_n =0
Nice work

============

 
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Sy123

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Re: HSC 2013 4U Marathon

Wow haha, not the solution I had in mind it can be done in 3 lines with cosine rule of non-right angled triangles. Either way good insight.

(I would love another one of those really hard questions even if I cant solve it I want to have a go)
 
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Re: HSC 2013 4U Marathon

Wow haha, not the solution I had in mind it can be done in 3 lines with cosine rule of non-right angled triangles. Either way good insight.

(I would love another one of those really hard questions even if I cant solve it I want to have a go)
haha, it was the first thing that came into my mind when I saw all those moduli



I think a similar question has been asked before, not sure lol
 

Sy123

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Re: HSC 2013 4U Marathon

haha, it was the first thing that came into my mind when I saw all those moduli



I think a similar question has been asked before, not sure lol
Yeah Realise asked something similar before and I answered it. I will post a question soon if I can make/find a good one.
 

Sy123

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Re: HSC 2013 4U Marathon

Here is an easy (?) result I just proved.



 

Sy123

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Re: HSC 2013 4U Marathon

not sure if this is exactly right lol. Are a, b and c are meant to be integers as well? Or just rational/real?

Yes, well divisibility implies integers. But you got it right anyway.

================





When I say cos k theta, I mean express cos^4 theta as cos theta, cos 2theta, cos 100 theta etc



(Still waiting for a good question)
 
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HeroicPandas

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Re: HSC 2013 4U Marathon

Yes, well divisibility implies integers. But you got it right anyway.

================





When I say cos k theta, I mean express cos^4 theta as cos theta, cos 2theta, etc



(Still waiting for a good question)
i) sub z and by DE MOIVRE'S...
LHS =z^n + z^-n

= cosn(theta) + isinn(theta) + cosn(theta) -isinn(theta) = 2cos ntheta as required

ii) (z+1/z)^4 = z^4 +1/z^4 + 4(z^2+1/z^2) +6

noting z^n +1/z^n = 2cosntheta

threfore, 16cos^4theta = 2cos4theta + 8cos2theta +6

cos^4theta = 1/8 cos4theta + 0.5 cos2theta + 3/8
 
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HeroicPandas

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Re: HSC 2013 4U Marathon

I) show that (a^2-1)(a^4-14a^2+1) =a^6 - 15a^4 + 15a^2 -1

II) by using de moivre's, show that: cot6theta = [(a^2-1)(a^4-14a^2+1)] / [2a (3a^4 -10a^2 +3)]

where a =cot(theta) and 6theta is not a multiple of PI

III)hence show that cot^2(pi/12) +cot^2(5pi/12) =14

IV)deduce that cot(pi/12) +tan (pi/12) =4
 
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Sy123

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Re: HSC 2013 4U Marathon

i) sub z and by DE MOIVRE'S...
LHS =z^n + z^-n

= cosn(theta) + isinn(theta) + cosn(theta) -isinn(theta) = 2cos ntheta as required

ii) (z+1/z)^4 = z^4 +1/z^4 + 6(z^2+1/z^2) +6

noting z^n +1/z^n = 2cosntheta

threfore, 16cos^4theta = 2cos4theta + 8cos2theta +6

cos^4theta = 1/8 cos4theta + 0.5 cos2theta + 3/8
Correct, good job.
 
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