在△ABC,已知tan[(A+B)/2]=sin C,求sin(C/2)的值请把过程列出来

用A＋B＝PAI－C 代入，然后右边用倍角公式展开，约掉sin（C/2），求出cos(C/2)，进而就可以算出sin(C/2)

你好，z1z1z1z1123
A+B=180° -C => tan[(A+B)/2]=tan[90° -C/2]=cot C/2 =(cosC/2) /(sinC/2)
sinC=2(sinC/2)(cosC/2) =>(cosC/2) /(sinC/2)=2(sinC/2)(cosC/2)
=>cosC/2 -2(sinC/2)^2*(cosC/2)=0 =>(cosC/2)*[1- 2(sinC/2)^2]=0
=> cosC/2=0 or (sinC/2)^2=1/2
=> C/2=0不符合=> sinC/2 =√2 /2
希望对你有帮助哈、

A+B=180° -C => tan[(A+B)/2]=tan[90° -C/2]=cot C/2 =(cosC/2) /(sinC/2)
sinC=2(sinC/2)(cosC/2) =>(cosC/2) /(sinC/2)=2(sinC/2)(cosC/2)
=>cosC/2 -2(sinC/2)^2*(cosC/2)=0 =>(cosC/2)*[1- 2(sinC/2)^2]=0
=> cosC/2=0 or (sinC/2)^2=1/2
=> C/2=0不合 => sinC/2 =√2 /2 ...ans

∵A+B=180-C.===>(A+B)/2=90-(C/2).===>tan[(A+B)/2]=tan[90-(C/2)]=con(C/2)=[cos(C/2)]/[sin(C/2)]=2sin(C/2)cos(C/2).===>sin²(C/2)=1/2.===>sin(C/2)=√2/2.