Polynomial bounds for sine and cosine

Step 1: $\sin x\le x$. Let $g_1(x)=x-\sin x$. Then $g_1'(x)=1-\cos x\ge0$, so $g_1$ is non-decreasing on $[0,+\infty)$ and $g_1(x)\ge g_1(0)=0$.

Step 2: $\cos x\ge1-\frac{x^2}2$. Let $g_2(x)=\cos x-1+\frac{x^2}2$. Then $g_2'(x)=x-\sin x=g_1(x)\ge0$ by Step 1, and $g_2(0)=0$, so $g_2\ge0$ on $[0,+\infty)$.

Step 3: $\sin x\ge x-\frac{x^3}6$. Let $g_3(x)=\sin x-x+\frac{x^3}6$. Then $g_3'(x)=\cos x-1+\frac{x^2}2=g_2(x)\ge0$ by Step 2, and $g_3(0)=0$, so $g_3\ge0$ on $[0,+\infty)$.

Application. Together they give $1-\frac{x^2}6\le\frac{\sin x}x\le1$ for $x>0$, which re-proves $\frac{\sin x}x\to1$ by squeezing.