---------------------------------------------------- Rappels de cours : Soient a et b des nombres réels. cos²(a)+sin²(a) = 1 cos(a-b) = cos(a)*cos(b)+sin(a)*sin(b) sin(a-b) = sin(a)*cos(b)-cos(a)*sin(b) ----------------------------------------------------
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Bonsoir,----------------------------------------------------
Rappels de cours :
Soient a et b des nombres réels.
cos²(a)+sin²(a) = 1
cos(a-b) = cos(a)*cos(b)+sin(a)*sin(b)
sin(a-b) = sin(a)*cos(b)-cos(a)*sin(b)
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D'après l'énoncé, cos(7π/12) = (√2-√6)/4
Or cos²(7π/12)+sin²(7π/12) = 1
D'où sin²(7π/12) = 1-cos²(7π/12)
D'où sin²(7π/12) = 1-((√2-√6)/4)² = 1-((√2-√6)²/16) = 1-((2-2√12+6)/16) = 1-((8-4√3)/16) = 1-((2-√3)/4) = (4-2+√3)/4 = (2+√3)/4
D'où sin(7π/12) = √((2+√3)/4) ou -√((2+√3)/4)
Or 7π/12 ∈ ]0;π[, d'où sin(7π/12) > 0
Donc sin(7π/12) = √((2+√3)/4) = (√(2+√3))/2
Ainsi :
cos(π/12) = cos((7π/12)-(π/2)) = cos(7π/12)*cos(π/2)+sin(7π/12)*sin(π/2) = ((√2-√6)/4)*0+sin(7π/12)*1 = sin(7π/12) = (√(2+√3))/2
sin(π/12) = sin((7π/12)-(π/2)) = sin(7π/12)*cos(π/2)-cos(7π/12)*sin(π/2) = sin(7π/12)*0-cos(7π/12)*1 = -cos(7π/12) = -(√2-√6)/4