一道函数数学题，在线等

1
cos(A+C)=cos(π-B)=-cosB=1/2
cosB=-1/2
B=2π/3.
A+C=π/3.
a=2sinA
则a/sinA=2
根据正弦定理有
b/sinB=c/sinC=a/sinA=2=2R,
则b=2sinB=2·sin(2π/3)=√3;
R=1;c=2sinC.
则由余弦定理,
=(4sin^2 A+4sin^2 C-3)/(2×2sinA×2sinC)
=[4(sin^2 A+sin^2 C)-3]/(8sinA·sinC)
=1/2
则4(sin^2 A+sin^2 C)-3=4sinA·sinC
4(sinA+sinC)^2=8sinA·sinC+3
根据和差化积以及积化和差公式,有
sinA+sinC=2sin[(A+C)/2]·cos[(A-C)/2]=2sin(π/6)·cos[(A-C)/2]=cos[(A-C)/2].
sinA·sinC=(1/2)[cos(A+C)-cos(A-C)]
=(1/2)·(1/2)-(1/2)cos(A-C)
=1/4-(1/2)cos(A-C)
则4·cos^2 [(A-C)/2]
=2-4cos(A-C)+3=5-4cos(A-C)=5-4{2·cos^2 [(A-C)/2]-1}=9-8cos^2 [(A-C)/2]
12cos^2 [(A-C)/2]=9;
cos [(A-C)/2]=√3/2.
(A-C)/2=±π/6
A-C=±π/3.
则cosC=cos[(A+C)/2-(A-C)/2]
=cos[(A+C)/2]·cos[(A-C)/2]-sin[(A+C)/2]·sin[(A-C)/2]
=(1/2)·(√3/2)±(√3/2)·(1/2)
=0或√3/2.
当cosC=0时,C=π/2;则B+C>π.显然不成立.
于是cosC=√3/2
(此时C=π/6;A