Question

sin(arctg2+arccos 4/5)

Answer 1

$sin(arctg2+arccos\frac{4}{5})\\sin(arctg2)cos(arccos\frac{4}{5})+cos(arctg2)sin(arccos\frac{4}{5})\\\\sin(arctg2)\\arctg2=x=tgx=2\ \textgreater \ 0\\sinx=\sqrt{\frac{1}{1+\frac{1}{tg^2x}}}=\sqrt{\frac{1}{1+\frac{1}{2^2}}}=\sqrt{\frac{1}{\frac{5}{4}}}\\sinx=\frac{2}{\sqrt5};$

$\\\\cos(arccos\frac{4}{5})=\frac{4}{5};\\\\\\cos(arctg2)\\arctg2=x=tgx=2\ \textgreater \ 0\\cosx=\sqrt{\frac{1}{1+tg^{2}x}}=\sqrt{\frac{1}{1+2^2}}\\cosx=\sqrt{\frac{1}{5}};\\\\sin(arccos\frac{4}{5})\\arccos\frac{4}{5}=x=cosx=\frac{4}{5}\ \textgreater \ 0\\sinx=\sqrt{1-cos^2x}=\sqrt{1-(\frac{4}{5})^2}\\sinx=\frac{3}{\sqrt5}$

$sin(arctg2)cos(arccos\frac{4}{5})+cos(arctg2)sin(arccos\frac{4}{5})\\ \frac{2}{\sqrt5}*\frac{4}{5}+\frac{1}{\sqrt5}*\frac{3}{\sqrt{5}}=\frac{8}{5\sqrt5}+\frac{3}{5}=\frac{8+3\sqrt5}{5\sqrt5}$