高中三角函数问题。急！谢谢！！在
1)1+tanA/tanB=2c/b
--->1+(sinA/cosA)/(sinB/cosB)=2(2RsinC)/(2RsinB)
--->1+(sinAcosB)/(cosAsinB)=2sinC/sinB
--->(sinAcosB+cosAsinB)/(cosAsinB)=2sinC/sinB
--->sin(A+B)/(cosAsinB)=2sin(A+B)/sinB
△ABC中sin(A+B)=sinC<>0,sinB<>0
--->2cosA=1
--->cosA=1/2
--->A=60°
2)m+n=(cosB,-1+2[cos(C/2)]^2)=(cosB,cosC)...全部
1)1+tanA/tanB=2c/b
--->1+(sinA/cosA)/(sinB/cosB)=2(2RsinC)/(2RsinB)
--->1+(sinAcosB)/(cosAsinB)=2sinC/sinB
--->(sinAcosB+cosAsinB)/(cosAsinB)=2sinC/sinB
--->sin(A+B)/(cosAsinB)=2sin(A+B)/sinB
△ABC中sin(A+B)=sinC<>0,sinB<>0
--->2cosA=1
--->cosA=1/2
--->A=60°
2)m+n=(cosB,-1+2[cos(C/2)]^2)=(cosB,cosC)
|m+n|=√[(cosB)^2+(cosC)^2]
=√[(1+cos2B)/2+(1+cos2C)/2]
=√[1+(cos2B+cos2C)/2]
=√[1+cos(B+C)cos(B-C)] 【和差化积】
=√[1-cosAcos(B-C)]
=√[1-(1/2)cos(B-C)]
当仅当B=C时cos(B-C)取得最大值1
--->1-cos(B-C)>=1-1=0
所以|m+n|有最小值0。
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