Question

【题目】已知函数f(x)=sin^2(wx+π/6)-cos^2(wx+π/6),(w>0)的最小正周期为2π,（

Answer 1

（1）f(x)=sin^2(wx+π/6)-cos^2(wx+π/6)
=sin^2(wx+π/6)-[1-sin^2(wx+π/6)]
=2sin^2(wx+π/6)-1
最小正周期=2π/W,所以W=1,所以f(x)=2sin^2(x+π/6)-1
（2）tanx=sinx/cosx=4/3,
（sinx）^2+（cosx）^2-1,且x∈（π,3π/2）,
sinx=-4/5,cosx=-3/5,
又f(x)=2sin^2(x+π/6)-1=2（sinxcosπ/6+cosxsinπ/6）^2-1
=（7+24√3）/50
“√”根号