已知函数f(x)=2sinx*sin(π/2+x)-2sin^2x+1若f(x0/2)=根2/3,x0∈(-π/4,π/4),求cos2x0

Question

已知函数f(x)=2sinx*sin(π/2+x)-2sin^2x+1
若f(x0/2)=根2 /3,x0∈(-π/4,π/4),求cos2x0

Answer 1

f(x)=2sinx*sin(π/2+x)-2sin^2x+1
＝2sinxcosx+cos2x
=sin2x+cos2x
=√2sin(2x+π/4)
因为f(x0/2)=根2 /3
所以
sin(x0+π/4)=1/3
cos2(x0+π/4)=1-2sin²(x0+π/4)=1-2×(1/3)²=7/9
即
sin2x0=-7/9
而 x0∈(-π/4,π/4)
2x0∈(-π/2,π/2)
即
2x0∈(-π/2,0)
又cos²2x0+sin²2x0=1
所以
cos2x0=4√2/9