...sin^2A-sin^2C)=(根号2*a-b)b,求三角形ABC面积的最大值

发布网友发布时间：2024-04-02 07:32

共2个回答

热心网友时间：2024-08-02 13:11

a/sinA=b/sinB=c/sinC=2R=>a/2R=sinA,b/2R=sinB,c/2R=sinC
带入4R^2(sin^2A-sin^2C)=(根号2*a-b)b=>a^2+b^2-c^2=√2ab由余弦定理得c^2=a^2+b^2-2abcosc=>√2ab=2abcosc=>cosc=√2/2=>C=45°S△ABC=absinC/2=2RsinA*2RsinB*sinC/2
=√2R^2(sinAsinB)=√2R^2[cos(A-B)-cos(A+B)]/2
=√2R^2[cos(A-B)-cos(π-c)]/2=√2R^2[cos(A-B)+cosc]/2=√2R^2[cos(A-B)+√2/2]/2
≤√2R^2[1+√2/2]/2=(1+√2)/2R^2

热心网友时间：2024-08-02 13:10

a/sinA=b/sinB=c/sinC=2R=2√2
=>a=2RsinA,b=2RsinB,c=2RsinC
2√2（sin�0�5A-sin�0�5C）=（a-b）sinB
=>4R�0�5（sin�0�5A-sin�0�5C）=2R（a-b）sinB
=>a�0�5-c�0�5=(a-b)b
=>(a�0�5 b�0�5-c�0�5)/2ab=1/2=cosC
=>C=60°

S△ABC=absinC/2=2RsinA*2RsinB*sinC/2
=√3(2sinAsinB)=√3[cos(A-B)-cos(A B)]
=√3[cos(A-B) 1/2]≤3√3/2