sinx^4是(sinx)^4?
y=sinx^4+cos²x
=sinx^4+1-sin²x
=sin²x(sin²x-1)+1
=-sin²xcos²x+1
=-1/4*sin²2x+1
=-1/4[sin²2x-1/2(sin²2x+cos²2x)+1/2]+1
= -1/4[1/2(sin²2x-cos²2x)+1/2]+1
=-1/8*[-cos4x]-1/8+1
=1/8cos4x+7/8
最小正周期为π/2