1、1/2[f²（π/4－x)＋f²(π/4＋x)]
=[cos²(π/4－x)＋cos²(π/4＋x)]/2
=[((cos(π/2-2x) 1)/2 ((cos(π/2 2x) 1)/2]/2
=[sin2x 1 (-sin2x 1)]/4
=2/4=1/2
f²（π/4－x）*f²（π/4＋x）=cos²(π/4－x)*cos²(π/4＋x)
=((cos(π/2-2x) 1)/2*((cos(π/2 2x) 1)/2=(1 sin2x)(1-sin2x)/4
=(1-sin²2x)/4=cos²2x/4