数学必修五数学题（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
(证明:作过三角形的一点如A外接圆的直径交圆于点D,连接BD,CD,根据同弦所对的圆周角相等=>角B等于角ADC,又角ADC是直角(直径所对的角是直径),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*根号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