bjr
on a sin(x + y) = cosx.siny + sinx.cosy
on exprime sin(x + y + π/2) de deux manières différentes
1) sin[(x + y) + π/2] = cos(x + y) (1) // [ sin(θ + π/2) = cosθ ]
2)
sin[x + (y + π/2)] = cosx . sin(y + π/2) + sinx . cos(y + π/2)
= cosx . cosy + sinx . (- siny) // [ cos(θ + π/2) = - sinθ ]
= cosx . cosy - sinx . siny (2)
(1) = (2)
