在三角形ABC中,角A、B、C满足sinC cosB=(2sinA-sinB)cosC(1)求角C的大小

sinC cosB=2sinAcosC-sinB cosC
sinC cosB+sinB cosC=2sinAcosC
sin(B+C)=2sinAcosC
由于三角形中A=180-(B+C)
sin(180-A)=2sinAcosC
sinA(1-2cosC)=0
2cosC=1
cosC=1/2
0y=2sinB^2-cos2A=1-cos2B-cos2A=1-cos2B-cos2(180-60-B)=1-cos2B-cos2(120-B)=1-cos2B-cos(240-2B)=1-cos2B-cos(120+B)=1-cos2B-cos120cos2B+sin120sin2B=1-cos2B+1/2cos2B+3^(1/2)/2sin2B=1-1/2cos2B+3^(1/2)/2sin2B=1+sin(2B-30)由于A +B ＝120，故0