Question

设A, B,C是三角形ABC的三个内角，求证：sin2A+sin2B+sin2C=4sinAsinBsinC

具体步骤，并解释原因

Answer 1

sin2A+sin2B+sin2C
=sin2A+sin2B+sin2(π-A-B)
=sin2A+sin2B+sin(2π-2A-2B)
=sin2A+sin2B+sin(-2A-2B)
=sin2A+sin2B-sin(2A+2B)
=sin2A+sin2B-sin2Acos2B-cos2Asin2B
=sin2A(1-cos2B)+sin2B(1-cos2A)
=2sin2A(sinB)^2+sin2B(sinA)^2
4sinAsinBsinC
=4sinAsinBsin(π-A-B)
=4sinAsinBsin(A+B)
=4sinAsinB(sinAcosB+cosAsinB)
=4sinAsinBsinAcosB+4sinAsinBcosAsinB
=2sin2A(sinB)^2+sin2B(sinA)^2
所以 成立