A maths－oob73zj92b的部落格

標題:

A maths

發問:

1 in triangle ABC, if cosA-cosB= sinC, prove that ABC is a right-angled triangle. 2 if A+B+C=180^。 and 2sinB=sinA+sinC, prove that tan(A/2)tan(C/2)=1/3 更新: -2cos(C/2)sin[(A-B)/2]=sin C -2cos(C/2)sin[(A-B)/2]=2sin(C/2)[cos(C/2)] why sinC becomes 2sin(C/2)[cos(C/2)] ?

最佳解答:

1 in triangle ABC, if cosA-cosB= sinC, prove that ABC is a right-angled triangle. Since cos A - cos B = sin C -2sin[(A+B)/2]sin[(A-B)/2]=sin C -2sin[(180-C)/2]sin[(A-B)/2]=sin C -2cos(C/2)sin[(A-B)/2]=sin C -2cos(C/2)sin[(A-B)/2]=2sin(C/2)[cos(C/2)] -sin[(A-B)/2]=sin(C/2) sin[(B-A)/2]=sin(C/2) (B-A)/2=C/2 B-A=C Because A + B + C = 180, A + B + B - A = 180, B=90 Prove that ABC is a right-angled triangle﹐B is a right-angled 2 if A+B+C=180^。 and 2sinB=sinA+sinC, prove that tan(A/2)tan(C/2)=1/3 2 sin B = sin A + sin C 2 sin(180-(A+C) )=2sin[(A+C)/2]cos[(A-C)/2] sin(A+C)=sin[(A+C)/2]cos[(A-C)/2] 2sin[(A+C)/2]cos[(A+C)/2]=sin[(A+C)/2]cos[(A-C)/2] 2cos[(A+C)/2]=cos[(A-C)/2] SO tan (A/2) tan (C/2) =[sin(A/2)sin(C/2)]/[cos(A/2)cos(C/2)] ={cos[(A-C)/2]-cos[(A+C)/2]}/{cos[(A-C)/2]+cos[(A+C)/2]} ={2cos[(A+C)/2]-cos[(A+C)/2]}/{2cos[(A+C)/2]+cos[(A+C)/2]} =1/3