在△ABC中,sinA,sinB,sinC成等差数列,求证:5cosA-4cosAcosC+5cosC=4
不用余弦定理,有什么简单方法
參考答案:2sin(A+C)=4sin(0.5A+0.5C)cos(0.5A+0.5C)
=2sinB=sinA+sinC=2sin(0.5A+0.5C)cos(0.5A-0.5C)
2cos(0.5A+0.5C)=cos(0.5A-0.5C)
5cosA-4cosAcosC+5cosC=5(cosA+cosC)-4cosAcosC
=10cos(0.5A+0.5C)cos(0.5A-0.5C)-2[cos(A+C)+cos(A-C)]
=10cos(0.5A+0.5C)cos(0.5A-0.5C)
-4cos(0.5A+0.5C)^2-4[cos(0.5A-0.5C)]^2+4
将2cos(0.5A+0.5C)=cos(0.5A-0.5C)代入
=20[cos(0.5A-0.5C)]^2-16[cos(0.5A-0.5C)]^2-4[cos(0.5A-0.5C)]^2+4
=4
熟练运用和差化积、积化和差公式求解