Question 1038719
sin^2 2x=(1-cos 4x)/2
sin^4 2x=(1-2cos 4x+cos^2 4x)/4
cos^2 4x=(1+cos 8x)/2
cos^2 4x/4=(1/8)+(cos 8x)/8
Therefore, sin^4 2x=(1/4)-(cos(4x))/2+[(1/8)+(cos 8x)/8)=(5/8)-[cos(4x)]/2+[cos(8x)/8],
and 2 sin^4(2x)=(5/4)-cos(4x)+[cos(8x)]/4