Question

求学霸解决这道题！

Answer 1

解，f(x)=cos^2x+2√3sin(兀/2+x)cos(3兀/2+x)
-sin^2x
=cos^2x-sin^2x+2√3cosxsinx
=cos2x+√3sin2x
=2sin(2x+兀/6)=2cos(2x-兀/3)
当x∈[0，兀/2]，则2x+兀/6∈[兀/6，7兀/6]
则f(x)最大=2，最小=2sin(7兀/6)=-1
f(a)=2cos(2a-兀/3)=6/5
tan^2(兀/6-a)=tan^2(a-兀/6)
=sin^2(a-兀/6)/cos^2(a-兀/6)
=(2sin^2(兀/6-a)/2cos(兀/6-a)sin(兀/6-a)]^2
=(1-cos(2a-兀/3)/sin(2a-兀/3)]^2
=(1-3/5)^2/(1-(3/5)^2)
=1/4

追问

学霸学霸，最大值2我带入π/6算出来是1