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+