作业帮在三角形ABC中,若sinA+sinB=sinC(cosA+cosB),c=1,则内切圆半径的取值范围,详细过程
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/19 10:37:29
在三角形ABC中,若sinA+sinB=sinC(cosA+cosB),c=1,则内切圆半径的取值范围,详细过程
在三角形ABC中,若sinA+sinB=sinC(cosA+cosB),c=1,则内切圆半径的取值范围,详细过程在三角形ABC中,若sinA+sinB=sinC(cosA+cosB),c=1,则内切圆半径的取值范围,详细过程
sinA+sinB=sinC(cosA+cosB)
2sin[(A+B)/2]cos[(A-B)/2]=2sin[(A+B)/2]2cos[(A+B)/2]2cos[(A+B)/2]cos[(A-B)/2]
1=2cos²[(A+B)/2]
cos(A+B)=0
A+B=C=90°
r[cot(A/2)+cot(B/2)]=c
r=sin(A/2)sin(B/2)/sin[(A+B)/2]
={cos[(A-B)/2]-cos[(A+B)/2]}/{2sin[(A+B)/2]}
=√2cos[(A-B)/2]/2-1/2
A+B=90°
-90°sin(A)+sin(B)=sin(A+B)(cos(A)+cos(B))
0=sin(A)cos(B)cos(B)+sin(B)cos(A)cos(A)-sin(A)-sin(B)+sin(A)cos(B)cos(A)+sin(B)cos(A)cos(B)
0=sin(A)cos(B)cos(A)+sin(B)cos(A)cos(B)-sin(B)sin(A)sin(A)-sin...全部展开
sin(A)+sin(B)=sin(A+B)(cos(A)+cos(B))
0=sin(A)cos(B)cos(B)+sin(B)cos(A)cos(A)-sin(A)-sin(B)+sin(A)cos(B)cos(A)+sin(B)cos(A)cos(B)
0=sin(A)cos(B)cos(A)+sin(B)cos(A)cos(B)-sin(B)sin(A)sin(A)-sin(A)sin(B)sin(B)
0=sin(B)cos(A+B)+sin(A)cos(A+B)
0=cos(A+B)(sin(B)+sin(A))
所以 cos(A+B)=0 即 角C为 pi/2
r=ab/(a+b+1)=sin(A)*cos(A)/(sin(A)+cos(A)+1)
=(1/2)*(sin(2A)/((2^0.5)*sin(A+pi/4)+1))
=0.5*(2*(sin(A+pi/4))^2-1)/((2^0.5)*sin(A+pi/4)+1))
=0.5*((2^0.5)*sin(A+pi/4)-1)
可见 当A=pi/4 时最大 r=0.5*((2^0.5)-1)
而r不为负，所以 最小 r=0收起