cos(π/6+α)=sin(π/3-α),sin(π/4+α)=cos(π/4-α)
所以,原式=sin(π/4-α)cos(π/3-α)-sin(π/3-α)cos(π/4-α)
=sin(π/4-α)cos(π/3-α)-cos(π/4-α)sin(π/3-α)
=sin[(π/4-α)-(π/3-α)]
=sin(-π/12)
=(√2-√6)/4