Question

已知△ABC的外接圆半径为R，且满足2R[(sinA)^2-(sinC)^2]=(√2a-b)sinB，求△ABC面积的最大值

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Answer 1

2R[(sinA)^2-(sinC)^2]=(√2a-b)sinB;两边同乘以2R得：[(2RsinA)^2-(2RsinC)^2]=(√2a-b)2RsinB
a^2-c^2=b(√2a-b); a^-c^2=√2ab-b^2; a^2+b^2-c^2=√2ab;
由余弦定理：cosC=(a^2+b^2-c^2)/(2ab)=√2ab/(2ab)=√2/2; 0<C<π
所以：C=π/4
S=(1/2)absinC=(1/2)4R^2sinAsinBsinC=(√2R^2)sinAsinB=(√2R^2)sinAsin(3π/4-A)
=(√2R^2)sinA[sin(3π/4)cosA)-cos(3π/4)sinA]=[sinAcosA+(sinA)^2]R^2
=[(1/2)sin2A+(1/2)(1-cos2A)]R^2=[(1/2)(sin2A-cos2A)+1/2]R^2
=[(√2/2)sin(2A-π/4)+1/2]R^2
0<A<3π/4; 0<2A<3π/2; -π/4<2A-π/4<5π/4
当2A-π/4=π/2; 即 A=3π/8时，S取到最大值(√2/2+1/2 )R^2