[微積] 不等式
if 0<x<π/2, prove 2/π<sinx/x<1
想法:微分均值定理
設f(x) = sinx ,
fx(a) = cosa = (sinx-sin0)/(x-0), where 0<a<x
因cosa<1 => sinx/x<1
想問左邊 2/π<sinx/x 怎麼証
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