sinus estimation by quadratic polynomial

Question

Prove that $\sin(\pi x)\ge \pi x(1-x)$ for $x\ge 0$.

I tried to use $\sin t\ge t-\frac{t^3}{6}$, but it is too weak here.

Answer 1

Consider $$f\left(x\right)=\sin\left(\pi x\right)-\pi x\left(1-x\right) $$ then $$f'\left(x\right)=\pi\left(\cos\left(\pi x\right)-1+2x\right) $$ and note that $$\cos\left(\pi x\right)-1+2x>0 $$ if $ 0<x<1/2 $ (this follows from the classical Jordan's inequality) or $x>1 $ then $x=0,1 $ are minimum for the function, so $$f\left(x\right)\geq\min\left\{ f\left(1\right),f\left(0\right)\right\} =0. $$

Answer 2

This estimate is symmetric about $x=\frac12$. Set $x=\frac12+t$, then the inequality becomes $$ \cos(πt)\ge \frac{\pi}4(1-4t^2) $$ Now start with $\cos(πt)\ge 1-\frac{\pi^2}2t^2$, ...