Bonjour,
cos(6π/5) = cos(2π/5 + 4π/5)
et cos(4π/5) = cos(2π/5 + 2π/5)
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
sin(a + b) = sin(a)cos(b) + sin(b)cos(a)
donc cos(2a) = cos²(a) - sin²(a) ou 2cos²(a) - 1 ou 1 - 2sin²(a)
et sin(2a) = 2sin(a)cos(a)
⇒ cos (2π/5 + 4π/5) = cos(2π/5)cos(4π/5) - sin(2π/5)sin(4π/5)
= cos(2π/5)(cos²(2π/5) - sin²(2π/5)) - 2sin²(2π/5)cos(2π/5)
etc...
cos(4π/5) = cos²(2π/5) - sin²(2π/5)
et sin(3π/10) = sin[(3π/5)/2]
si on pose a = 3π/5, sin(3π/10) = sin(a/2)
Or sin²(a/2) + cos²(a/2) = 1
⇒ sin²(a/2) = 1 - cos²(a/2) = [cos(2xa/2) + 1]/2 = (cos(a) + 1)/2
donc sin(a/2) = √[(1 + cos(a))/2]
⇒ sin(3π/10) = √[(1 + cos(3π/5))/2]
