三角函数最值问题已知x,y,z为实数,求：f(x,y,z)=[sin(x-y)]^2+[sin(y-z)]^2+[sin(z-x)]^2的最大值.

Question

三角函数最值问题
已知x,y,z为实数,求：f(x,y,z)=[sin(x-y)]^2+[sin(y-z)]^2+[sin(z-x)]^2的最大值.

Answer 1

sin^2(x-y)+sin^2(y-z)+sin^2(z-x)
=[1-cos2(x-y)+1-cos2(y-z)+1-cos2(z-x)]/2
=3/2-[(cos2xcos2y+sin2xsin2y)+(cos2ycos2z+sin2ysin2z)+(cos2zcos2x+sin2zsin2x)]/2
=3/2-[(2cos2xcos2y+2cos2ycos2z+2cos2zcos2x)+(2sin2xsin2y+2sin2ysin2z+2sin2zsin2x)+cos^2(2x)+sin^2(2x)-1+cos^2(2y)+sin^2(2y)-1+cos^2(2z)+sin^2(2z)-1]/4(这步非常关键)
=3/2-[(sin2x+sin2y+sin2z)^2+(cos2x+cos2y+cos2z)^2-3]/4
≤3/2+3/4=9/4
当x=π/3,y=2π/3,z=π时,sin2x+sin2y+sin2z)^2=(cos2x+cos2y+cos2z)^2=0
上式可以取到等号.故最大值是9/4