Chord and Lipschitz bounds for sine

On $[0,\pi \mathrm{/}2]$ define $h(t)=\sin t-t\cos t$. By [F1]–[F3] and the product rule, $h$ is differentiable and continuous, with $h(0)=0$ and $h^{\mathrm{\prime }}(t)=t\sin t\ge 0$ on $(0,\pi \mathrm{/}2)$. For $0<t\le \pi \mathrm{/}2$, the mean value theorem on $[0,t]$ gives $h(t)-h(0)=h^{\mathrm{\prime }}(c)t\ge 0$ for some $c\in (0,t)$; at $t=0$ equality holds.

If $u=0$, then $\mathrm{\mid }\sin u\mathrm{\mid }=0=\mathrm{\mid }u\mathrm{\mid }$ by [F1]. For $0<\mathrm{\mid }u\mathrm{\mid }\le \pi \mathrm{/}2$, sine is continuous on $[0,\mathrm{\mid }u\mathrm{\mid }]$ and differentiable on its interior by [F1] and [F3], so the mean value theorem applies on this nondegenerate interval. Its derivative has modulus at most $1$ by [F1] and [F5], so $\mathrm{\mid }\sin \mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$; parity gives $\mathrm{\mid }\sin u\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$.

For $0<t<\pi \mathrm{/}2$, put $r(t)=\sin t\mathrm{/}t$. The quotient rule and step 1.1 give $r^{\mathrm{\prime }}(t)=(t\cos t-\sin t)\mathrm{/}{t}^2=-h(t)\mathrm{/}{t}^2\le 0$. Applying the mean value theorem to $r$ on $[t,\pi \mathrm{/}2]$ yields $r(t)\ge r(\pi \mathrm{/}2)=2\mathrm{/}\pi$ by [F1] and the quarter-turn value in [F1]; the endpoint $t=\pi \mathrm{/}2$ is equality, and $t=0$ is immediate. Parity in [F5] extends the lower bound to negative $u$.

If $u\mathrm{\ne }v$, apply the mean value theorem to $\sin$ on the closed interval with endpoints $u,v$. For some interior $c$, $\mathrm{\mid }\sin u-\sin v\mathrm{\mid }=\mathrm{\mid }\cos c\mathrm{\mid }\mathrm{\mid }u-v\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$ by [F1], [F4], and [F5]; if $u=v$, both sides are zero.