在△ABC中,已知内角A=π/3,边BC=2√3,设内角B=x,周长为y,求y的解析式和y的最大值

Question

在△ABC中,已知内角A=π/3,边BC=2√3,设内角B=x,周长为y ,求y的解析式和y的最大值

Answer 1

由正弦定理,得
BC/sinA=AC/sinB=AB/sinC
从而 AC=BC*sinB/sina=2√3*sinX/sinπ/3=2√3*sinX/(√3/2)=4sinX
AB=BC*sinC/sina=2√3*sin(π-B-A)/sinπ/3=2√3*sin(π-x-π/3)/(√3/2)=4sin(2π/3-x)
∴y=AC+AB+BC
=4sinX+4sin(2π/3-x)+2√3
=4(sinX+sin(2π/3-x)+2√3
=4*2*sin((2π/3/2)*cos((2π/3-2x)/2)+2√3 { sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2] }
=8*sinπ/3*cos(π/3-x)+2√3
=8*√3/2*cos(π/3-x)+2√3
=4√3*cos(π/3-x)+2√3
则 y的解析式是 y=4√3*cos(π/3-x)+2√3
∵cos(π/3-x)的最大值为1
∴y的最大值=4√3*1+2√3=6√3