0≤x≤π2, then show that x≥sin⁡x

Let f(x)=x−sin⁡xNow f1(x)=1−cos⁡x≥0 in 0≤x≤π2f1(x)≥0So, f(x) is increasing functionf(x)≥0⇒x−sin⁡x≥0 in x∈0,π2x≥sin⁡x

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$0 \leq x \leq \frac{π}{2}, then show that x \geq \sin x$

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Let $f (x) = x - \sin x$

Now $f^{1} (x) = 1 - \cos x \geq 0 in 0 \leq x \leq \frac{π}{2}$

$f^{1} (x) \geq 0$

So, $f (x)$ is increasing function

$f (x) \geq 0 \Rightarrow x - \sin x \geq 0 in x \in (0, \frac{π}{2})$

$x \geq \sin x$

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