已知函数f(x)=2sinxsin(π/2+x)-2sin²x+1求最小正周期及单调递增区间若f(x①/2)=根号2/3,x①∈(-π/4,π/4),求cos2x①的值

Question

已知函数f(x)=2sinxsin(π/2+x)-2sin²x+1求最小正周期及单调递增区间
若f(x①/2)=根号2/3,x①∈(-π/4,π/4),求cos2x①的值

Answer 1

f(x)=2sinxsin(π/2+x)-2sin²x+1
=2sinxcosx+1-2sin²x
=sin2x+cos2x
=√2sin(2x+π/4)
所以最小正周期为 2π/2=π
f(x①/2)=√2/3
√2sin(2x①+π/4)=√2/3
得 sin(2x①+π/4)=1/3
sin(2x①)+cos(2x①)=√2/3
平方得 1+2sin(2x①)cos(2x①)=2/9
得 2sin(2x①)cos(2x①)=-7/9
cos(2x①)-sin(2x①)=4/3
相加得
cos(2x①)=(4+√2)/6