Question

已知函数f(x)=2sinxcosx+cos2x(x

已知函数f(x)=2sinxcosx+cos2x(x

Answer 1

f(x)=sin2x+cos2x
=√2sin(2x+π/4)
所以T=2π/2=π
最大值=√2

f(θ+π/8)=√2sin(2θ+π/4+π/4)
=√2cos2θ
=√2/3
cos2θ=1/3
θ锐角则sin2θ>0
sin²2θ+cos²2θ=1
sin2θ=2√2/3
tan2θ=2√2

Answer 2

解：
f(x)=2sinxcosx+cos(2x)
=sin(2x)+cos(2x)
=√2sin(2x+π/4)
最小正周期Tmin=2π/2=π
当sin(2x+π/4)=1时，有最大值f(x)max=√2
当sin(2x+π/4)=-1时，有最小值f(x)min=-√2