Mathc initiation/a0026
Vérifier quelques propriétés mathématiques de trigonométrie
Vérifions si : cos(x) - cos(y) = -2 sin( (x+y)/2 ) sin( (x-y)/2 ) = +2 sin( (x+y)/2 ) sin( (y-x)/2 )
posons :
cos(x) - cos(y) = -2 sin( (x+y)/2 ) sin( (x-y)/2 )
Soit :
= -2 sin( x/2 + y/2 ) sin( x/2 - y/2 )
Nous avons vu que :
sin(x+y) = cos(x) sin(y) + sin(x) cos(y)
sin(x-y) = sin(x) cos(y) - cos(x) sin(y)
donc
cos(x) + cos(y) = -2 [sin( x/2+y/2 )]
[sin( x/2-y/2 )]
= -2 [cos(x/2) sin(y/2) + sin(x/2) cos(y/2)]
[sin(x/2) cos(y/2) - cos(x/2) sin(y/2)]
= -2 [ cos(x/2) sin(y/2) sin(x/2) cos(y/2)
- cos(x/2)**2 sin(y/2)**2
+ sin(x/2)**2 cos(y/2)**2
- sin(x/2) cos(y/2) cos(x/2) sin(y/2)]
= -2 [- cos(x/2)**2 sin(y/2)**2
+ sin(x/2)**2 cos(y/2)**2
+ cos(x/2) sin(y/2) sin(x/2) cos(y/2)
- sin(x/2) cos(y/2) cos(x/2) sin(y/2)]
= -2 [- cos(x/2)**2 sin(y/2)**2
+ sin(x/2)**2 cos(y/2)**2 ]
sin(x/2) = sqrt((1-cos(x))/2)
cos(x/2) = sqrt((1+cos(x))/2)
= -2 [- sqrt((1+cos(x))/2)**2 sqrt((1-cos(y))/2)**2
+ sqrt((1-cos(x))/2)**2 sqrt((1+cos(y))/2)**2 ]
= -2 [- (1+cos(x))/2 (1-cos(y))/2
+ (1-cos(x))/2 (1+cos(y))/2 ]
= -2 [- (1+cos(x)) (1-cos(y)) /4
+ (1-cos(x)) (1+cos(y)) /4 ]
= -1/2 [- (1+cos(x)) (1-cos(y))
+ (1-cos(x)) (1+cos(y)) ]
= -1/2 [-( 1-cos(y)+cos(x)-cos(x)cos(y))
+ 1+cos(y)-cos(x)-cos(x)cos(y)]
= -1/2 [- 1+cos(y)-cos(x)+cos(x)cos(y)
+ 1+cos(y)-cos(x)-cos(x)cos(y)]
= -1/2 [ +cos(y)-cos(x)
+cos(y)-cos(x)]
= -1/2 [ +2cos(y)-2cos(x)]
= -2/2 [ cos(y)-cos(x) ]
= cos(x) - cos(y)