Question

已知f(x)=sin(wx+θ)cos(wx+θ)+2cos∧2(wx+θ)-1,其中0≤θ≤π/2,w＞0

Answer 1

1、

f(x)=1/2*sin(2wx+2θ)+cos(2wx+2θ)

=√(1/4+1)sin(2wx+2θ+z)

tanz=1/(1/2)=2

T=2π/(2w)=2π

w=1/2

2、

f(x)=√(1/4+1)sin(x+2θ+z)

对称轴是x=0

sinx对称轴是x=kπ+π/2

把x=0代入

0+2θ+z=kπ+π/2

2θ=kπ+π/2-z

tan2θ=tan(kπ+π/2-z)=tan(π/2-z)=cotz=1/tanz=1/2

2tanθ/(1-tan²θ)=1/2

tan²θ+4tanθ-1=0

tanθ>0

所以tanθ=2+√5

θ=arctan(2+√5)