f(x)=√[(√3)²+1²]*sin(wx+q-z)
=2sin(wx+q-z)
tanz=1/√3
所以z=π/6
f(x)=2sin(wx+q-π/6)
sin的對(duì)稱軸是取最值的地方,所以兩相鄰對(duì)稱軸的距離是半個(gè)周期
所以T=π
所以T=2π/π=2
偶函數(shù)
f(-x)=2sin(-2x+q-π/6)=f(x)=2sin(2x+q-π/6)
所以-2x+q-π/6=2kπ+2x+q-π/6
或-2x+q-π/6=2kπ+π-(2x+q-π/6)
-2x+q-π/6=2kπ+2x+q-π/6
這個(gè)不是恒等式,舍去
-2x+q-π/6=2kπ+π-(2x+q-π/6)
2q=2kπ+4π/3
q=kπ+2π/3
所以q=2π/3
f(x)=2sin(2x+2π/3)
f(π/8)=2sin(π/4+2π/3)=2sinπ/4cos2π/3+2cosπ/4sin2π/3=(√6-√2)/4