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
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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.
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.
![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](/uploads/image/z/618099-51-9.jpg?t=%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E6%9C%80%E5%80%BC%E9%97%AE%E9%A2%98%E5%B7%B2%E7%9F%A5x%2Cy%2Cz%E4%B8%BA%E5%AE%9E%E6%95%B0%2C%E6%B1%82%EF%BC%9Af%28x%2Cy%2Cz%29%3D%5Bsin%28x-y%29%5D%5E2%2B%5Bsin%28y-z%29%5D%5E2%2B%5Bsin)
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
=[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
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