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sin()sin{()}  ≦ x  sincoscossin ≦ sincossin  sin()sin()sin{()} ≦ yx  y ≦ cos() ≦  ≦  10,02 102 sin()sin()sin{()} ≦ yx  ≦≦≦≦  xy 10,0,01,cos()2

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Academic year: 2021

シェア "sin()sin{()}  ≦ x  sincoscossin ≦ sincossin  sin()sin()sin{()} ≦ yx  y ≦ cos() ≦  ≦  10,02 102 sin()sin()sin{()} ≦ yx  ≦≦≦≦  xy 10,0,01,cos()2"

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[ 東京工業大学 1973 年 ３ ]

0 , 0 1 , 0 1, cos( )

2 x y

 

  

 

≦ ≦ ≦ ≦ であるとき，不等式

sin(



)≦ysin(



)xsin{(

  

 ) }

が成り立つならば，x1であることを示せ。

0 , 0 1

 

  

2

≦ より 1

0≦



 2

なので sin

 

 ≧0 …①

cos( )

y≦



の両辺に sin

 

 をかけると，①より ysin

 

 ≦cos() sin

 

 …②

よって，sin(



)≦ysin(



)xsin{(

  

 ) }

 

cos() sin  xsin{(   ) }

≦

移項して sin() cos( ) sin

 

 ≦xsin{(   ) } …③

また，0≦

 

⇔   

 

≦0 ⇔ 0 

  

≦ と 1

0 

2 より

0 ( ) 1

   

2

  ≦  であるから sin(

  

 ) 0 …④ さらに 1

0 

2 より 1

0



 2

であるから sin(



)0 …⑤

したがって，④，⑤より

 

0sin{(   ) }sin   sin



cos



cos



sin



≦sin



cos



sin



…⑥

③，⑥より sin(

  

 ) ≦xsin{(

  

 ) } が成り立つので x1 である。

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