1、解：y=[(sinx)^2]^2=[(1颍骈城茇-cos2x)/2]^2=(1/4)(1-2cos2x+（cos2x）^2)=(1/4)-(1/2)cos2x+(1/4)*(1/2)(1+cos4旌忭檀挢x)=(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像粜杵泳)*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)
