Question

f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))

Answer 1

f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))

=(cos2x)^2/((sinπ/4cosx+cosπ/4sinx)(sinπ/4cosx-cosπ/4sinx))

=(cos2x)^2/(cosx^2/2-sinx^2/2)

=2cos2x

1.f(-11π/12)=2cos(-11π/6)=√3

2.g(x)=cos2x+sin2x=√2sin(2x+π/4)

当X∈[0,π/4)时,2x+π/4∈[π/4,3π/4).所以最大值为√2,最小值为1