TEST N°1(préparation)

Test n°1

http://maths-akir.midiblogs.com/media/00/02/90708694.pdf

Pour l'exercice 3 :

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)

= cos(p/18) / sin(p/9).sin(2p/9) = 1/ [2.sin(p/18).sin(2p/9)]

d'où

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 puisque sin(4p/9) = cos(p/2 - 4p/9) = cos(p/18)

Salut Xploze
Xploze à écrit (plus claire)