Question

已知圆O的半径为R,它的内接△满足2R（sin²A－sin²C）=（√2a－b）sinB,

求△ABC面积的最大值。

Answer 1

解：2R（sin²A－sin²C）=（√2a－b）sinB
2R（a²－c²)/4R^2=（√2a－b）b/2R
a^2+b^2-c^2=√2ab
cosC=(a^2+b^2-c^2)/(2ab)=√2/2
C=Л/4,A+B=3Л/4
S=1/2absinC=√2ab/4=√2R^2sinAsinB=-√2R^2[cos(A+B)-cos(A-B)]/2
=√2R^2[cos(A-B)+cosC]/2
=√2R^2[cos(A-B)+√2/2]/2
所以cos(A-B)=1时，S有最大值=√2R^2+1/2