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设实数x,y,z满足0<x<y<z<π2,证明:π2+2sinxcosy+2sinycosz>sin2x+sin2y+sin2z.
题目详情
设实数x,y,z满足0<x<y<z<
,证明:
+2sinxcosy+2sinycosz>sin2x+sin2y+sin2z.
π |
2 |
π |
2 |
▼优质解答
答案和解析
证明:由于sin2x+sin2y+sin2z-2sinxcosy-2sinycosz
=
[(sin2x+sin2y)+(sin2y+sin2z)+(sin2z+sin2x)]-2sinxcosy-2sinycosz
≤sin(x+y)cos(x-y)+sin(y+z)cos(y-z)+sin(z+x)cos(z-x)-2sinxcosycos(x-y)-2sinycoszcos(y-z)
=sin(y-x)cos(x-y)+sin(z-y)cos(y-z)+sin(z+x)cos(z-x)
=
sin(2y-2x)+
sin(2z-2y)+sin(z+x)cos(z-x)
=sin(z-x)cos(2y-x-z)+sin(z+x)cos(z-x)
≤sin(z-x)+cos(z-x)≤
<
故
+2sinxcosy+2sinycosz>sin2x+sin2y+sin2z.
=
1 |
2 |
≤sin(x+y)cos(x-y)+sin(y+z)cos(y-z)+sin(z+x)cos(z-x)-2sinxcosycos(x-y)-2sinycoszcos(y-z)
=sin(y-x)cos(x-y)+sin(z-y)cos(y-z)+sin(z+x)cos(z-x)
=
1 |
2 |
1 |
2 |
=sin(z-x)cos(2y-x-z)+sin(z+x)cos(z-x)
≤sin(z-x)+cos(z-x)≤
2 |
π |
2 |
故
π |
2 |
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