Bonsoir,
je décompose tout :
1) cos(17pi/2 - x) = cos(pi/2+8pi-x) = cos(pi/2-x+2x4pi) = cos(pi/2-x)=sin(x)
sin(x+3pi) = sin(x+pi+2pi)=sin(x+pi)=-sin(x)
cos(3pi/2-x) = -sin(x)
sin(13pi-x) = sin(-x+13pi) = sin(-x+pi-12pi)=sin(-x+pi+2x6pi)=sin(-x+pi)=sin(x)
ce qui donne
A(x) = sin(x)-(-sin(x))-sin(x)-sin(x)
A(x) = sin(x) + sin(x) - sin(x) - sin(x)
A(x) = 2sin(x)
2) sin(x)^2 - sni(x)^4 = sin(x)^2(1 - sin(x)^2) = cos(x)^2 x sin(x)^2
cos(x)^2 - cos(x)^4 = cos(x)^2(1 - cos(x)^2) = cos(x)^2 x sin(x)^2
cos(x)^2 x sin(x)^2
B(x) = -------------------------- = 1
cos(x)^2 x sin(x)^2
3) cos( pi / 5) ^2 = cos(pi/2 - 3pi/10) ^2 = sin(3pi/10) ^2 utilise cos(A-B)
ce qui donne
sin(3pi/10)^2 + cos(3pi/10)^2 = 1
car sin(a)^2 + cos(a)^2 = 1 vient de pythagore ---> Apprendre par coeur
5) Rappel tan(x) = sin(x) / cos(x)
ce qui donne :
sin(x) ^2 / cos(x) ^2 sin(x) ^2 / cos(x) ^2
--------------------------- = -----------------------------------------
1 + (sin(x)^2 / cos(x)^2) (cos(x)^2 + sin(x)^2) / cos(x)^2
sin(x) ^2 cos(x) ^2 sin(x) ^2
------------- x ---------------------- = ---------------------
cos(x) ^2 cos(x)^2 + sin(x)^2 cos(x)^2 + sin(x)^2
et cos(x)^2 + sin(x)^2 = 1
donc c'est bien égal à sin(x)^2
j'en ai marre
A+
