分类：教育/科学-理工学科-数学
问题：高中数学化简sin^a+COS^aCOS（π/3+a）
化简sin^a+OS^aCOS（π/3+a）-sin^（π/6-a）
回答：∵(sinα)^2-[sin(π/6-α)]^2
=(1-cos2α)/2-[1-cos(π/3-2α)]/2
=(1/2)cos(π/3-2α)-(1/2)cos2α
且cosαcos(π/3+α)
={cos[(π/3+α)+α]+cos[(π/3+α)-α]}/2
=1/4+(1/2)cos(π/3+2α)
∴(sinα)^2+cosαcos(π/3+α)-[sin(π/6-α)]^2
=[(1/2)cos(π/3-2α)-(1/2)cos2α]+[1/4+(1/2)cos(π/3+2α)]
=1/4-(1/2)cos2α+(1/2)[cos(π/3+2α)+cos(π/3-2α)]
=1/4-(1/2)cos2α+cos(π/3)cos2α
=1/4-(1/2)cos2α+(1/2)cos2α
=1/4
=常数，
因此，原式值与α无关。