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For 0<xlt=pi/2,s howt h a tx-(x^3)/6<sin...

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Let f(x) = sin x-x
`therefore f(x) =cosx-1=-(1-cosx)=-2sin^(2)x//2lt0`
Thus f(x) is a decreasing function
Now `x gt 0 `
or `f(x) lt f(x) or sinx -x lt 0`
Now `let g(x) =x-(x^(3))/(6) - sinx`
`therefore g(x) =1-(x^(2))/(2)-cosx`
To find sign of g(x) we consider
Thus `phi` (x) is a decreasing function .Therefore
since `x gt0` we get `g(x) ltg(x)`
or `x-(x^(3))/(6)-sinxlt0`
or `x-(x^(3))/(6)ltsinx`
combininig equation (1) and (2) we get
`x-(-x^(3))/(6)lt sinx ltx`
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