1、解：
y=[(sinx)^2]^2
=[(1-cos2x)/2]^2
=(1/4)(1-2cos2x+（cos2x）^2)
=(1/4)-(1/2)cos2x+(1/4)*(1/2)(1+cos4x)
=(1/4)-(1/2)cos2x+(1/8)+(1/8)cos4x
=(1/8)cos4x-(1/2)cos2x+(3/8)
2、y’=-(1/2)sin4x+sin2x
=2^(2*1-3) *cos(4x+π/2)-2^(1-1)*cos(2x+π/2)
y’’=-2cos4x+2cos2x
=2^(2*2-3) *cos(4x+2*π/2)-2^(2-1)*cos(2x+2*π/2)
y’’’=8sin4x-4sin2x
=2^(2*3-3) *cos(4x+3*π/2)-2^(3-1)*cos(2x+3*π/2)
y(4)=32cos4x-8cos2x
=2^(2*4-3) *cos(4x+4*π/2)-2^(4-1)*cos(2x+4*π/2)
y(5)=-128sin4x+16sin2x
=2^(2*5-3) *cos(4x+5*π/2)-2^(5-1)*cos(2x+5*π/2)
3、所以：y(n)= 2^(2*n-3)*cos(4x+nπ/2)-2^(n-1)*cos(2x+nπ/2)
