Помогите с решением добрые люди

Решите задачу:

$arccos(sin4)=?\\\\sin4\approx sin229,2^\circ \in 3\; chetverti\; \; \Rightarrow \;\; sin4=sin(\pi +(4-\pi ))=-sin(4-\pi )\\\\sin(4-\pi )=cos\Big (\frac{\pi}{2}-(4-\pi )\Big )\\\\arccos(sin4)=arccos\Big (-sin(4-\pi )\Big )=\pi -arccos\Big (sin(4-\pi )\Big )=\\\\=\pi -arccos\Big (cos(\frac{\pi}{2}-(4-\pi))\Big )=\pi -\Big (\frac{\pi }{2}-(4-\pi )\Big )=\pi -(\frac{3\pi}{2}-4)=\\\\=4-\frac{\pi}{2}$

$ili\\\\\quad arccos(sin4)=arccos(cos\underbrace {(\frac{\pi}{2}-4)}_{\in 3 chetvert})=arccos(cos\underbrace {(4-\frac{\pi}{2})}_{\in 2 chetvert})=4-\frac{\pi}{2}\\\\\\\star \; \; arccos(cosa)=a\; ,\; esli\; \; 0\leq a\leq \pi \; \; \star$