Question

这两道题过程

Answer 1

15)
f(x)=2sinwxcos(wx+π/3)
=2sinwx(1/2coswx+√3/2sinwx)
=1/2sin2wx+√3/2*2(sinwx)^2
=1/2sin2wx+√3/2(1-cos2wx)
=1/2sin2wx-√3/2cos2wx+√3/2
=sin(2wx-π/3)+√3/2
T=2π/2w=π
w=1
f(x)=sin(2x-π/3)+√3/2
x在[-π/6,π/2]

2x-π/3x在[-2π/3,2π/3]
sin(2x-π/3)在[-2π/3,2π/3]上值域为:[-√3/2,1]
f(x)=sin(2x-π/3)+√3/2,在[-2π/3,2π/3]上值域为:[0,(2+√3)/2]
最小值=0,最大值=(2+√3)/2
16)
Sn=2n^2-n ,a1=S1=1
an=Sn-S(n-1)=2n^2-2(n-1)^2-n+(n-1)=4n-3
an=4n-3
bn=(-1)^nan=(-1)^n*(4n-3)
Tn=-1+5-9+13-17...+(4n-3)=-(1+16n-15)/2+(5+16n-3)/2
=-(8n-7)+(8n+1)
=8