Prove that, $\cos x\ge1-{2x\over\pi}\ \forall x\in[0,{\pi\over2}]$

Question

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 1

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.

Answer 2

Simple: the function $\cos x$ is concave on the interval $[0,\pi/2]$, hence any chord on this interval is under the arc it subtends. In particular, the chord joining the $x$-intercept $(\pi/2,0)$ and the $y$-intercept $(0,1)$.

Now, an equation of the line through $(a,0)$ and $(0,b)$ is $\dfrac xa+\dfrac yb=1$, so here we have the equation $$\frac{2x}\pi +\frac y1=1\iff y=1-\frac{2x}\pi,$$ and the chord being under the curve arc translates as $$1-\frac{2x}\pi \le\cos x.$$