Question

sin(arctan 2 - arccos(5/7)) вычислить​

Answer 1

$\displaystyle\star\star\star\\\\\displaystyle\sin(a-b)=\sin a\cos b-\cos a\sin b\\\\\displaystyle\sin(\mathrm{arctg}\,a)=\cfrac{a}{\sqrt{1+a^2}}\\\\\\\displaystyle\cos(\mathrm{arccos}\,a)=a\\\\\displaystyle\cos(\mathrm{arctg}\,a)=\cfrac{1}{\sqrt{1+x^2}}\\\\\\\displaystyle\sin(\arccos\,a)=\sqrt{1-x^2}\\\\\displaystyle\star\star\,\star$

$\displaystyle\sin\bigg(\mathrm{\arctg2}-\arccos\cfrac{5}{7}\bigg)=\\\\\\=\sin(\mathrm{arctg}\,2)\cos\bigg(\arccos\cfrac{5}{7}\bigg)-\cos(\mathrm{arctg}\,2)\sin\bigg(\arccos\cfrac{5}{7}\bigg)=\\\\\\=\cfrac{2}{\sqrt{1+2^2}}\cdot\cfrac{5}{7}-\cfrac{1}{\sqrt{1+2^2}}\cdot\sqrt{1-\bigg(\cfrac{5}{7}\bigg)^2}=\cfrac{10}{7\sqrt{1+4}}-\cfrac{\sqrt{1-\cfrac{25}{49}}}{\sqrt{1+4}}=$

$=\cfrac{10}{7\sqrt5}-\dfrac{\sqrt{\cfrac{24}{49}}}{\sqrt5}=\cfrac{10\sqrt5}{35}-\cfrac{\cfrac{\sqrt{24}}{7}}{\sqrt5}=\cfrac{2\sqrt5}{7}-\cfrac{\sqrt{24}}{7\sqrt5}=\cfrac{2\sqrt5}{7}-\cfrac{2\sqrt{30}}{35}$