Bonjour,
A(x) = cos(3x) + sin(2x)/2 - 2cos³(x) + 2cos(x)
1) A(-π/6) = cos(-3π/6) + sin(-2π/6)/2 - 2cos³(-π/6) + 2cos(-π/6)
= cos(-π/2) + sin(-π/3)/2 - 2cos³(-π/6) + 2cos(-π/6)
cos(-π/2) = 0
sin(-π/3) = -√(3)/2
cos(-π/6) = √(3)/2
⇒ A(-π/6) = -√(3)/4 - 2(√(3)/2)³+ √(3)
= -√(3)/4 - 3√(3)/4 + 4√(3)/4
= 0
A(π/4) = cos(3π/4) + sin(2π/4)/2 - 2cos³(π/4) + 2cos(π/4)
cos(3π/4) = -√(2)/2
sin(2π/4) = sin(π/2) = 1
cos(π/4) = √(2)/2
⇒ A(π/4) = -√(2)/2 + 1/2 - 2(√(2)/2)³ + √(2)
= √(2)/2 + 1/2 - √(2)/2
= 1/2
2) cos(3x) = cos(2x + x)
= cos(2x)cos(x) - sin(2x)sin(x)
= [cos²(x) - sin²(x)]cos(x) - [sin(x)cos(x) + sin(x)cos(x)]sin(x)
= cos(x)(1 - 2sin²x) - 2sin²(x)cos(x)
= [1 - 4sin²(x)]cos(x)
A(x) = cos(3x) + sin(2x)/2 - 2cos³(x) + 2cos(x)
= [1 - 4sin²(x)]cos(x) + 2sin(x)cos(x)/2 - 2[1 - sin²(x)]cos(x) + 2cos(x)
= cos(x)[1 - 4sin²(x) + sin(x) -2 + 2sin²(x) + 2]
= cos(x)[-2sin²(x) + sin(x) + 1]
On retrouve :
A(-π/6) = √(3)/2[-2x1/4 - 1/2 + 1] = 0
A(π/4) = √(2)/2[-2x2/4 + √(2)/2 + 1] = 1/2
