高考解三角函数题

cosA+cosB=cosC………………(2)

从(2) 2-(1) 2

cos2A+cos2B+2(cosa cosb-Sina sinb)= cos2C

so 2cos(a+b)cos(a-b)+2cos(a+b)= cos2c.............(3)

它也是由(1) 2+(2) 2导出的。

2+2(cosAcosB+sinAsinB)=1

因此，COSA COSB+Sina sinb = COS(A-B)=-1/2.......................................(4).

所以从(3)

2 cos(A+B)*(-1/2)+2 cos(A+B)= cos2C

所以cos2c = cos(a+b)= cosa cos b-Sina sinb...........................................................(5)

然后从(4)-(5)得到

2sinAsinB=-1/2-cos2C

因此

正弦平方A+正弦平方B+正弦平方c

=(sina+sinb)^2-2sinasinb+(sinc)^2

=2(sinC)^2-(-1/2-cos2C)

=2(sinC)^2+1/2+cos2C

=2(sinc)^2+1/2+1-2(sinc)^2

=3/2