salam : p = pi = 3,1415....
l'outil : formule du double et de transformation(somme en produit) , angles complémentaires
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1) 1/sin(p/9) + 1/sin(2p/9) = [sin(2p/9) + sin(p/9)] / sin(p/9).sin(2p/9)
= 2.sin(3p/18).cos(p/18) / sin(p/9).sin(2p/9)
or sin(3p/18)=sin(p/6)=1/2
on continue = cos(p/18) / sin(p/9).sin(2p/9) = 1/ [2.sin(p/18).sin(2p/9)]
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1/sin(p/9) + 1/sin(2p/9) - 1/ sin(4p/9)
=1/[2.sin(p/18).sin(2p/9)] - 1/ [2.sin(2p/9).cos(2p/9)]
= [1/2sin(2p/9)].[ 1/sin(p/18) - 1/ cos(2p/9) ]
= [1 / 2sin(2p/9) ].[ ( cos(2p/9) - sin(p/18) ) / sin(p/18)cos(2p/9) ]
= [ 1 / 2sin(2p/9) ].[ (cos(2p/9) - cos(4p/9) ) / sin(p/18)cos(2p/9) ]
= [ 1 / 2sin(2p/9) ].[-2sin(p/3)sin(-p/9) / sin(p/18)cos(2p/9) ]
=[V3.2cos(p/18) / sin(4p/9) ]
=2V3
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c'est un peu délicat ( patience)
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