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prove trig (1 Viewer)

cxlxoxk

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Rkiuyto said:
prove that

sin 2*sin 18*sin 22*sin 38*sin 42*sin 58*sin 62*sin 78*sin 82 = 0.00019398655 (9 d.p)

sqrt (5 - 1) / 1 024 = 0.001953125 (9 d.p)

it doesn't? they are very close though

or am i just bad at trig cos i did general maths.
 
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Timothy.Siu

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cxlxoxk said:
sin 2*sin 18*sin 22*sin 38*sin 42*sin 58*sin 62*sin 78*sin 82 = 0.00019398655 (9 d.p)

sqrt (5 - 1) / 1 024 = 0.001953125 (9 d.p)

it doesn't? they are very close though

or am i just bad at trig cos i did general maths.
dude, its coz ur calculator sux,
u used radians.... when its in degrees,
and anyways the question is (root5-1)/1024 not sqrt(5-1)/1024...
 
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kaz1

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Let sin2=sin@

LHS=sin@*sin9@*sin11@*sin19@*sin21@*sin29@*sin31@*sin39@*sin41@

I can only think of De Moivre's theorem after this point.
 

dp624

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something to do with complex numbers
products of roots etc

i'll think more later...lol
 

gurmies

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^ I place my bets on neither. This question is similar to the format of mathematics olympiad/competition questions.
 

shaon0

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Ok guys,
I'll start it off.
Let x=2 degrees.
sin(x)*sin(20-x)sin(20+x)sin(40-x)sin(40+x)sin(60-x)sin(60+x)sin(80-x)sin(80+x)
Let y=20 degrees.
sin(x)*sin(y-x)sin(y+x)sin(2y-x)sin(2y+x)sin(3y-x)sin(3y+x)sin(4y-x)sin(4y+x)
= sin(x)*(((sin(y))^2(cos(x))^2)-(sin(x))^2(cos(y))^2))(((sin(2y))^2(cos(x))^2)-(sin(x))^2(cos(2y))^2))(((sin(3y))^2(cos(x))^2)-(sin(x))^2(cos(3y))^2))(((sin(4y))^2(cos(x))^2)-(sin(x))^2(cos(4y))^2))
 
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