Question

在三角形ABC的外接圆半径为R,且2R(sin^2A-sin^2C)=(根号2倍a-b)sinB,求三角形ABC面积的最

Answer 1

2R[(sinA)^2-(sinC)^2]=2R[a^2/(2R)^2-c^2/(2R)^2]=[1/(2R)](a^2-c^2)(√2a-b)sinB=[1/(2R)](√2ab-b^2)由题意知,[1/(2R)](a^2-c^2)=[1/(2R)](√2ab-b^2)即a^2-c^2=√2ab-b^2cosC=(a^2+b^2-c^2)/(2ab)=√2/2,则C=π/...