數(shù)學必修五數(shù)學題（正弦定理、余弦定理）

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

a/sinA=b/sinB=c/sinC=2R
(證明:作過三角形的一點如A外接圓的直徑交圓于點D,連接BD,CD,根據(jù)同弦所對的圓周角相等=>角B等于角ADC,又角ADC是直角(直徑所對的角是直徑),在Rt△ADC中,有b/sin角ADC=b/sinB=2R)
推出sinA=a/2R,sinB=b/2R.sinC=c/2R代入2R(sin^2 A-sin^2 C)=(a*根號2-b)sinB
化解得a^2-c^2=ab√2-b^2 (1)
即(a^2+b^2-c^2)/2ab=√2/2=cosC,推出C=45°
所以S△ABC=absinC/2=ab√2/4
=(2sinAsinB)√2/8=[cos(A-B)-cos(A+B)]√2/8
=[cos(A-B)+1/2]√2/8≤3√2/16