Question

已知函数f(x)=（4cos^4x-2cos2x-1）/[sin(π/4+x)·sin(π/4-x)]

(1)求f（-11/12π）的值
（2）当x∈[0,π/4）时，求g(x)=1/2f(x)+sin2x的最大和最小值

Answer 1

f(x)=（4cos^4x-2cos2x-1）/[sin(π/4+x)·sin(π/4-x)]
=[(1+cos2x)*2-2cos2x-1]/[(-1/2)(cosπ/2-cos2x)]
=2(cos2x)^2/cos2x=2cos2x
1)f(（-11/12)π）=2cos(-11π/6)=2cos(11π/6)=2cos(2π-π/6)=2cos(π/6)=根号3
2)g(x)=1/2f(x)+sin2x=cos2x+sin2x=根号(2)sin(2x+π/4),
当2x+π/4=π/2,x=π/8时，g(x)最大=根号2,当x=0时，g(x)最小=g(0)=1.