在三角形ABC中,tan[(A+B)/2]=2sinC.(1)求角C的大小（2）若A*B=1,求三角形ABC周长的取值范围.（角C的值不是90度而是60度,取值范围也不是)2

tan(A+B)/2=sinC
tan(180-C)/2=sinC
tan(90-(C/2))=sinC
cot(C/2)=sinC

cotc/2=cos(c/2)/sin(c/2)
sinc=2sin(c/2)cos(c/2)
所以有cos(c/2)/sin(c/2)=2sin(c/2)cos(c/2)
1＝2sin²(c/2)
sin²(c/2)=1/2
sin(c/2)=（根号2）/2
C/2=45
C=90
AB=1,那么AC=AB*cosA,BC=ABsinA
那么周长L=AB+AC+BC=1+sinA+cosA=1+根号2 sin(A+45)
由于0所以,根号2/2即:1+根号2*根号2/2那么周长的范围是:2

tan(A B)/2=sinC tan(180-C)/2=sinC tan(90-(C/2))=sinC cot(C/2)=sinC cotc/2=cos(c/2)/sin(c/2) sinc=2sin(c/2)cos(c/2)

一楼你个烧饼 抄错题了
tan(A+B)/2=2sinC
tan(180-C)/2=2sinC
tan(90-(C/2))=2sinC
cot(C/2)=2sinC
cotc/2=cos(c/2)/sin(c/2)
sinc=2sin(c/2)cos(c/2)
所以有cos(c/2)/sin(c/2)=4sin(c/2)cos(c/2)
1＝4sin²(c/2)
sin²(c/2)=1/4
sin(c/2)=1/2
C/2=30
C=60
一楼你个烧饼 抄错题了
tan(A+B)/2=2sinC
tan(180-C)/2=2sinC
tan(90-(C/2))=2sinC
cot(C/2)=2sinC
cotc/2=cos(c/2)/sin(c/2)
sinc=2sin(c/2)cos(c/2)
所以有cos(c/2)/sin(c/2)=4sin(c/2)cos(c/2)
1＝4sin²(c/2)
sin²(c/2)=1/4
sin(c/2)=1/2
C/2=30
C=60
AB=1,那么AC=AB*sin(120-A),BC=ABsinA
那么周长L=AB+AC+BC= [1+sinA+sin(120-A)] 由于0