real analysis - Prove that, $cos xge1-{2xoverpi} forall xin[0,{piover2}]$

By drawing the graph of $y=\cos x$ and $y=1-{2x\over\pi}$ on $[0,{\pi\over2}]$, it looks obvious due to convex nature of cosine function, which follows-

But how to prove it rigorously, using Taylor Series expansion or integration or by any other analysis.
Can anybody prove it?
Thanks for assistance in advance.

Answer

Let $f(x)=\cos(x)+\dfrac2\pi x$. Then $f'(x)=\dfrac2\pi-\sin(x)$ and, since $\dfrac2\pi\in(0,1)$ and $\sin$ is increasing in $\left[0,\frac\pi2\right]$, there is one and only one $x_0\in\left[0,\frac\pi2\right]$ such that $\sin(x_0)=\dfrac2\pi$. So, $f$ is increasing on $[0,x_0]$ and decreasing on $\left[x_0,\frac\pi2\right]$. But then, since $f(0)=f\left(\frac\pi2\right)=1$, you have $$\left(\forall x\in\left[0,\frac\pi2\right]\right):f(x)\geqslant1,$$ which is exactly what you want to prove.