已知x属于[45°,90°],且f(x)=2sin^2(x+45°)－根号3cos2x,求f(x)的最值

题目

已知x属于[45°,90°],且f(x)=2sin^2(x+45°)－根号3cos2x,求f(x)的最值

答案

f(x)=2sin²(x+45)-√3cos2x=[1-cos2(x+45)]-√3cos2x=1-cos(2x+90)-√3cos2x=1-sin(-2x)-√3cos2x=1+sin2x-√3cos2x=1+2[(1/2)sin2x-(√3/2)cos2x]=1+2[cos60sin2x-sin60cos2x]=1+2sin(2x-60)因为45≤x≤90所以...