Докажите, что если a,b,y - углы треугольника, то выполняется равенство: a) 4cos a/2*cos b/2*cos y/2 = sin a + sin b + sin y б) 4sin a/2*sin b/2*cos y/2 = sin a + sin b - sin y
А) 4cos a/2*cos b/2*cos y/2 = sin a + sin b + sin y --- 4cos α/2*cos β/2*cosγ/2 = 2(cos(α+β)/2 +cos(α-β)/2)*cosγ/2 = 2cos(α+β)/2*cosγ/2 +2cosγ/2 *cos(α-β)/2= cos(α+β+γ)/2 +cos(α+β-γ)/2+cos(α+γ-β)/2 +cos(γ+β-α)/2 = cosπ/2 +cos(α+β+γ -2γ)/2+cos(α+β+γ-2β)/2 +cos(β+γ+α-2α)/2= cos(π -2γ)/2+cos(π-2β)/2 +cos(π-α)/2= cos(π/2 -γ)+cos(π/2-β) +cos(π/2-α) = sinα +sinβ+sinγ. ---------- б) 4sin(α/2)*sin(β/2)*cos(γ/2) = sin α + sin β - sin γ --- sin α + sin β - sinγ =2sin((α+c)/2)*cos((α-β)/2) -sin(π-(α+β))= 2sin((α+β)/2)*cos((α-β)/2) -sin(α+β)= 2sin((α+β)/2)*cos((α-β)/2) -sin2*((α+β)/2)= 2sin((α+β)/2)*cos((α-β)/2) -2sin((α+β)/2)*cos((α+β)/2) = 2sin((α+β)/2)*(cos((α-β)/2) -cos((α+β)/2) )= 2sin((π-γ)/2) *(-2sin(α/2)*sin(-β/2) =2sin(π/2-γ/2) *2sin(α/2)*sin(β/2)= 2cos(γ/2) *2sin(α/2)*sin(β/2) =4sin(α/2)*sin(β/2)*cos(γ/2) .