Question

Using the relation $2(1-\cos x)<x^2,x=0$ or prove that $\sin \left(\tan x\right)\ge x,\forall ϵ[0,\pi /4]$

Answer 1

We know $sin\left(x\right)\ge tan\left(x\right)$ and $xϵ$ [0,2] or cos x$\ge \frac{1}{1+x^2}$

Given $2(1-cosx)<x^{2}x\ne 0$

$\Rightarrow (1-cosx)<\frac{x^2}{2}$

$cosx>(1-\frac{x^2}{2})$

Multiply both sides $(1+x^2)$ or $xϵ[0,1]$

$(1+x^2)cosx>(1-\frac{x^2}{2})$

Hence for any $zϵ[0,1]$ we have $cos\left(z\right)\ge \frac{1}{1+z^2}$ &

$\forall uϵ$ [0,1] sin (u) = ${\int }_0^{u}coszdz\ge {\int }_0^4\frac{dz}{1+z^2}$ are tan(u)

so by setting u = tan(x)

$xϵ$[0,$\frac{\pi }{4}],sin(tan\left(x\right))\ge x.$