Question

三角形ABC中，求证sin²A/2+sin²B/2+sin²C/2=1-2sinA/2*sinB/2*sinC/2

Answer 1

证明：∵cosA+cosB+cosC
=2cos[(A+B)/2]cos[(A-B)/2]+1-2sin²(C/2)
=2sin(C/2)cos[(A-B)/2]+1-2sin²(C/2)
=1+2sin(C/2){cos[(A-B)/2]-sin(C/2)}
=1+2sin(C/2){cos[(A-B)/2]-cos[(A+B)/2]}
=1+2sin(C/2)×2sin(A/2)sin(B/2)
=1+4sin(A/2)sin(B/2)sin(C/2)
又∵cosA+cosB+cosC
=1-2sin²(A/2)+1-2sin²(B/2)+1-2sin²(C/2)
=3-2[sin²(A/2)+sin²(B/2)+sin²(C/2)]
∴1+4sin(A/2)sin(B/2)sin(C/2)
=3-2[sin²(A/2)+sin²(B/2)+sin²(C/2)]
整理即得：
sin²(A/2)+sin²(B/2)+sin²(C/2)=1-2sin(A/2)sin(B/2)sin(C/2)