已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π）,且F（a/4)=根号2/4,求C

已知函数FX=(2COS^2X-1)SIN2X+1/2COS4X,若a=(π/2,π）,且F（a/4)=根号2/4,求COSa
F(x)=(2COS^2X-1)SIN2X+1/2COS4X=COS2XSIN2X+1/2COS4X
=1/2SIN4X+1/2COS4X=√2/2sin(4x+π/4)
F(a/4)=√2/2sin(a+π/4)=√2/4==>sin(a+π/4)=1/2==>a+π/4=5π/6==>a=7π/12
所以cosa=cos(π/2+π/12)=-sin(π/12)=-(√6-√2)/4

解
f(x)=(2cos²-1)sin2x+1/2cos4x
=cos2xsin2x+1/2cos4x
=1/2sin4x+1/2cos4x
=√2/2sin(4x+π/4)
∴
T=2π/4=π/2为其最小周期
最大值为：√2/2
f(a)=√2/2sin(4a+π/4)=√2/2
∴4a+π/4=π/2+2kπ