Trigonometry

Basic recall

1. ${\sin }^{\prime }(x)=\cos (x)$
2. ${\cos }^{\prime }(x)=-\sin (x)$
3. $\sin (0)=0,\cos (0)=1$
4. $\sin (\frac{\pi }{2})=1,\cos (\frac{\pi }{2})=0$
5. $\sin (-x)=-\sin (x)$

Def 41.1

$$\pi =\sup \{x\mid \forall 0<y<x,\cos (y)=\infty \}$$

Lemma 41.2

$$|\sin (x)|\le |x|$$

proved by Thm 32.1 Fundamental Theorem and $(\forall t)\cos (t)\le 1$ with

$$\sin (x)-\sin (0)={\int }_0^x{\sin }^{\prime }(t)dt={\int }_0^x\cos (t)dt\le x\le |x|$$

Then

$$-\sin (x)=\sin (-x)\le |x|$$

$\square$

Remark 41.3

$$1-\cos (x)={\int }_0^x\sin (t)dt$$