f(x)=√3sinx+3cosx,三角形ABC中,f(A)=2√3,BC=1,E为BC中点,则AE的最大值为

Question

f(x)=√3sinx+3cosx,三角形ABC中,f(A)=2√3,BC=1,E为BC中点,则AE的最大值为

Answer 1

答：f(A)=√3sinA+3cosA=2√3
sinA/2+√3cosA/2=1
sin(A+60°）=1
A+60°=90°
A=30°
根据正弦定理：AB/sinC=AC/sinB=BC/sinA=1/sin30°=2
根据余弦定理：
AE^2=AB^2+BE^2-2BE*ABcosB
=AB^2-ABcosB+1/4
=4(sinC)^2-2sinCcosB+1/4
=4(sinC)^2-2sinCcos(150°-C）+1/4
=3(sinC)^2+√3sinCcosC+1/4
=3(1-cos2C)/2+√3sin2C/2+1/4
=7/4+√3(sin2C/2-√3cos2C/2)
=7/4+√3sin(2C-60°）
故当2C-60°=90°即：B=C=75°时,
AE最大值为：√（7/4+√3)