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Topic: Trig identity involving pi/11 (Read 12850 times) |
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SWF
Uberpuzzler
Posts: 879
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Re: prove
« Reply #25 on: Mar 6th, 2007, 8:34pm » |
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This riddle is on the brink of slipping into oblivion due to its non-descriptive title. How about changing the title from "prove" to something that might help in finding this again- maybe "Trig identity involving pi/11".
It wasn't easy, but I found a way do it using basic trig identities:
(1) 2*sin(A)*cos(B) = sin(A+B) + sin(A-B)
(2) sin(2A)=2*sin(A)*cos(B) and cos(2A)=1-2*sin(A)^2 = 2*cos(A)^2-1
(3) sin(3A)=sin(A)*(3-4*sin(A)^2) and cos(3A)=cos(A)*(1-4*sin(A)^2)
(4) sin(A)^2 + cos(A)^2 = 1
Start with 2*sin(11x)*cos(x):
2*sin(11x)*cos(x) = sin(12x) + sin(10x) (from (1) with A=11x, B=x)
= sin(12x) + 2*sin(6x)*cos(4x) - sin(2x) (from a rearrangment of (1) with A=6x, B=4x)
= 2*sin(6x)*cos(6x) + 2*sin(6x)*(1 - 2*sin(2x)^2) - sin(2x) (from (2) )
= 2*sin(6x)*(2*cos(3x)^2 - 1 + 1) - 4*sin(6x)*sin(2x)^2 -sin(2x) (from (2) )
= sin(2x)*[2*(3-4*sin(2x)^2)*2*cos(3x)^2 - 4*sin(6x)*sin(2x) - 1 ] (from (3) and factoring out sin(2x)
= sin(2x)*[12*cos(3x)^2 - 16*sin(2x)^2*cos(3x)^2 - 8*sin(3x)*cos(3x)*sin(2x) - cos(3x)^2 -sin(3x)^2] (from (2), (4) and expanding)
= 2*sin(x)*cos(x)*[11*cos(3x)^2 - 16*sin(2x)^2*cos(3x)^2 - 8*sin(3x)*cos(3x)*sin(2x) -sin(3x)^2] (from (2) )
Factoring and dividing through by 2*cos(x) leaves:
sin(11x) = sin(x)*[ 11*cos(3x)^2 - (4*sin(2x)*cos(3x) + sin(3x) )^2 ]
If sin(11x) is zero and sin(x) is not zero (i.e. x=N*pi/11, N not a multiple of 11), the term in brackets is zero and dividing by cos(3x)^2 leaves:
11 = ( 4*sin(2x) + tan(3x) ) ^2
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Michael Dagg
Senior Riddler
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Posts: 500
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Re: Trig identity involving pi/11
« Reply #26 on: Mar 7th, 2007, 4:20pm » |
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KUDOS -- that's great SWF!
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| « Last Edit: Mar 7th, 2007, 4:43pm by Michael Dagg » |
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Michael Dagg
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JiNbOtAk
Uberpuzzler
Hana Hana No Mi
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Re: Trig identity involving pi/11
« Reply #27 on: Mar 8th, 2007, 4:58pm » |
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I can't believe I'm saying this, but I could actually understand SWF's approach ! 
Attempting it is a different story though, nice job SWF.
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Quis custodiet ipsos custodes?
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Michael Dagg
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Posts: 500
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Re: Trig identity involving pi/11
« Reply #28 on: Mar 9th, 2007, 9:21pm » |
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Of course, I hope others see this too as it is elegant, clear,
concise and constructive proof. There are some cagey
chess moves there but those are SWF secrets -- Gauss once
said, responding to admiration of the ingenuity of his proofs,
"that he'd simply removed the scaffolding (as is done with
impressive buildings) which had let him do it all."
(Try using SWF result to get also tan(4 pi/11) + 4 sin(pi/11) = sqrt(11) ).
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| « Last Edit: Mar 9th, 2007, 10:13pm by Michael Dagg » |
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Michael Dagg
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Michael Dagg
Senior Riddler
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Posts: 500
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Re: Trig identity involving pi/11
« Reply #29 on: Nov 7th, 2008, 4:07pm » |
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FYI there is recent paper by V.H. Moll in the
College Mathematics Journal, Vol. 39 No. 5.,
pp. 395-399 regarding trig identties involving
sqrt 11 that is neat.
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Regards,
Michael Dagg
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