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

sin(π/2+x) = cosx
sin(x/2) = √[(1 - cosx)/2]-> -2sin²x+1 = cos(2x)
f(x) = 2sinxsin(π/2+x)-2sin²x+1
= 2sinxcosx + cox(2x)
= sin(2x) + sin(2x + π/2)
= sin(2x + π/2) + sin(2x)
= 2sin[(2x + 2x + π/2)/2]cos[(2x + π/2 - 2x)/2]
= 2sin(2x + π/4)cos(π/4)
= √2sin(2x + π/4)
最小正周期为π， 在 -π/2 < 2x + π/4 < π/2时，f(x)单调递增。 即 -3π/8 < x < π/8
考虑到周期性，单调递增区间为(-3π/8 + kπ, π/8 + kπ), 这里k为整数。

x①∈(-π/4,π/4)时, cos2x① > 0
f(x/2) = √2sin(x + π/4) = √2/3
sin(x + π/4) = 1/3
sinxcos(π/4) + cosxsin(π/4) = 1/3
sinx + cosx = √2/3
平方：sin²x + cos²x + 2sinxcosx = 2/9
1 + sin(2x) = 2/9
sin(2x) = -7/9
cos(2x) = √[1 - sin²(2x)]
= √[1 - (-7/9)²]
= 4√2/9

Answer 2

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