Mathc initiation/a0023
Apparence
Vérifier quelques propriétés mathématiques de trigonométrie
Vérifions si : sin(x) + sin(y) = 2 sin( (x+y)/2 ) cos( (x-y)/2
posons :
sin(x) + sin(y) = 2 sin( (x+y)/2 ) cos( (x-y)/2 )
Soit :
= 2 sin( x/2 + y/2 ) cos( x/2 - y/2 )
Nous avons vu que :
sin(x+y) = cos(x) sin(y) + sin(x) cos(y)
cos(x-y) = cos(x) cos(y) + sin(x) sin(y)
donc
sin(x) + sin(y) = 2 [sin( x/2 + y/2 )]
[cos( x/2 - y/2 )]
= 2 [cos(x/2) sin(y/2) + sin(x/2) cos(y/2)]
[cos(x/2) cos(y/2) + sin(x/2) sin(y/2)]
= 2 [cos(x/2)**2 sin(y/2)cos(y/2) +
sin(y/2)**2 sin(x/2)cos(x/2) +
cos(y/2)**2 sin(x/2)cos(x/2) +
sin(x/2)**2 sin(y/2)cos(y/2)]
= 2 [cos(x/2)**2 sin(y/2)cos(y/2) +
sin(x/2)**2 sin(y/2)cos(y/2) +
cos(y/2)**2 sin(x/2)cos(x/2) +
sin(y/2)**2 sin(x/2)cos(x/2) ]
cos(x)**2+sin(x)**2=1
= 2[ (cos(x/2)**2+sin(x/2)**2) sin(y/2)cos(y/2)) +
(cos(y/2)**2+sin(y/2)**2) sin(x/2)cos(x/2) ) ]
= 2[ sin(y/2)cos(y/2) + sin(x/2)cos(x/2) ]
sin(x/2) = sqrt((1-cos(x))/2)
cos(x/2) = sqrt((1+cos(x))/2)
= 2[ sqrt((1-cos(y))/2) sqrt((1+cos(y))/2) +
sqrt((1-cos(x))/2) sqrt((1+cos(x))/2) ]
= 2[ sqrt( (1-cos(y))/2 (1+cos(y))/2 ) + sqrt(x) sqrt(y) = sqrt(xy)
sqrt( (1-cos(x))/2 (1+cos(x))/2 ) ]
= 2[ sqrt( (1-cos(y)) (1+cos(y))/4) +
sqrt( (1-cos(x)) (1+cos(x))/4) ]
= sqrt( (1-cos(y)) (1+cos(y)) ) + (x-y)(x+y) = x**2-y**2
sqrt( (1-cos(x)) (1+cos(x)) )
= sqrt( (1-cos(y)**2) ) + sin(x)**2+cos(x)**2=1
sqrt( (1-cos(x)**2) )
1-cos(x)**2 = sin(x)**2
= sqrt(sin(y)**2) +
sqrt(sin(x)**2)
= sin(y) + sin(x)