Réponse :
ex2 écrire plus simplement les expressions suivantes:
A = sin(π-x) + 3sin(x + π) - sin x + sin(- x) - 2sin(3π - x)
= sin x - 3 sin x - sin x - sin x - 2sin((π - x) + 2π)
= sin x - 3 sin x - sin x - sin x - 2sin x
A = - 6 sin x
or sin((π - x) + 2π) = sin(π-x)cos 2π + cos(π-x)sin2π
= sin(π - x) car cos 2π = 1 et sin 2π = 0
= sin x
B = cos(2π - x) + cos(x + π) + 5cos x + cos (- x) + 7(- 3π - x)
cos(2π - x) = cos((π - x) + π) = cos(π-x)cosπ - sin(π-x)sin π
= - cos x * (- 1) - sin(π - x)*0
= cos x
cos(x + π) = - cos x
cos (- x) = cos x
cos (- 3π - x) = cos((2π - x) - 5π) = cos(2π-x)cos 5π + sin(2π - x) sin 5π
or cos 5π = - 1 et sin 5π = 0
cos(- 3π - x) = cos(2π - x) = cos((π - x) + π) = cos(π - x)cosπ - sin(π-x)sinπ
cosπ = - 1 et sinπ = 0
cos(-3π - x) = cos(π - x) = - cos x
on obtient : B = cos(2π - x) + cos(x + π) + 5cos x + cos (- x) + 7(- 3π - x)
= cos x - cos x + 5 cos x + cos x - 7 cos x
= - cos x
B = - cos x
C = sin(π/2 - x) + 4 sin(x + π/2) + sin(3π/2 - x)
= cos x + 4 cos x - cos x = 4 cos x
C = 4 cos x
sin(3π/2 - x) = sin((π/2 - x) + π) = sin(π/2 - x)cosπ + cos(π/2 - x)sin π
cos π = - 1 et sin π = 0. Or sin(π/ - x) cos π = - cos x
Explications étape par étape