sin^4x+cos^4x-3/4
=sin^4x+2sin^2xcos^2x+cos^4x-2sin^2xcos^2x-3/4
=(sin^2x+cos^2)^2-2sin^2xcos^2x-3/4
=1-2sin^2xcos^2x-3/4
=1/4-2sin^2xcos^2x
=1/4-1/2(sin2x)^2
=1/4-1/2(sin2x)^2
=1/4-1/2[(1-cos4x)/2]
=1/4-1/4+(cos4x)/4
=(cos4x)/4
你再把x的范围带进去
-π/6