Нужна помощь с задачей номер 45

Решите задачу:

$\star \; \; \alpha =arcsin \frac{4}{5}\quad \Rightarrow \quad sin \alpha = \frac{4}{5}\; \; \Rightarrow \\\\cos \alpha =\sqrt{1-sin^2 \alpha }=\sqrt{1-\frac{16}{25}}= \frac{3}{5}\; \; \Rightarrow \; \; \alpha =arccos\frac{3}{5}\; \; \star \\\\\star \star \; \; arccosx+arccos y=arccos (xy-\sqrt{1-x^2}\cdot \sqrt{1-y^2})\; ,\\\\ esli\; x+y\geq 0\; \; \; \star \star \\\\\\arcsin \frac{4}{5} +arccos\frac{1}{\sqrt{50}} =arccos \frac{3}{5}+arccos \frac{1}{5\sqrt2} =\\\\=arccos\Big (\frac{3}{5}\cdot \frac{1}{5\sqrt2}-\sqrt{1-\frac{9}{25}}\cdot \sqrt{1-\frac{1}{50}}\Big )=$

$=arccos \Big ( \frac{3\sqrt2}{50}-\sqrt{\frac{16}{25}}\cdot \sqrt{\frac{49}{50} }\Big )=arccos\Big ( \frac{3\sqrt2}{50}-\frac{4}{5}\cdot \frac{7}{5\sqrt2}\Big )=\\\\=arccos \Big ( \frac{3\sqrt2}{50}-\frac{28\sqrt2}{50}\Big )=arccos\Big (\frac{-25\sqrt2}{50}\Big )=arccos (- \frac{\sqrt2}{2} )=\\\\=\pi -arccos\frac{\sqrt2}{2}=\pi -\frac{\pi}{4}=\frac{3\pi }{4}$