题目

求一道数学题（应该是相似的吧）
在三角形ABC中,∠A：∠B：∠C=1：2：4,
求证：1/AB+1/AC=1/BC

提问时间：2021-03-02

答案

∠A：∠B：∠C=1：2：4,
A = π/7
B = 2π/7
C = 4π/7
1/sinB + 1/sinC
= (sinB+sinC)/(sinBsinC)
= 4sin[(B+C)/2]cos[(B-C)/2]/[cos(B-C)-cos(B+C)]
= 4sin(3π/7)cos(π/7)/[cos(2π/7)-cos(6π/7)]
= 4sin(3π/7)cos(π/7)/[cos(2π/7)+cos(π/7)]
= 2sin(3π/7)cos(π/7)/[cos(3π/14)cos(π/14)]
= 2sin(3π/7)cos(π/7)/[cos((3π/14)sin(3π/7)]
= 2cos(π/7)/[cos((3π/14)]
= 2cos(π/7)sin(π/7)/[sin(π/7)cos(3π/14)]
= sin(2π/7)/[sin(π/7)cos(3π/14)]
= cos(3π/14)/[sin(π/7)cos(3π/14)]
= 1/sin(π/7)
= 1/sinA
1/sinB +1/sinC = 1/sinA
由正弦定理
1/AB+1/AC=1/BC