Sin(arctg(1/2)-arcctg(-sqrt(3))

[latex]sin(arctg\frac{1}{2}-arcctg(-\sqrt{3}))=\\ sin(arctg\frac{1}{2}+\frac{\pi}{3})=\\ sin(arctg\frac{1}{2})cos\frac{\pi}{3}+cos(arctg\frac{1}{2})sin\frac{\pi}{3}=\\ [/latex]   выразим синус через тангенс  [latex]sin(arctg\frac{1}{2})cos\frac{\pi}{3}+cos(arctg\frac{1}{2})sin\frac{\pi}{3}\\\\ 1)sin(arctg\frac{1}{2})=\frac{2tg\frac{arctg\frac{1}{2}}{2}}{1+tg^2\frac{arctg\frac{1}{2}}{2}}\\ tg\frac{arctg\frac{1}{2}}{2}=\frac{tg(arctg\frac{1}{2})}{1+\sqrt{1+tg(arctg\frac{1}{2})^2}}=\frac{\frac{1}{2}}{1+\sqrt{1+\frac{1}{4}}}=\frac{1}{2+\sqrt{5}}\\ sin(arctg\frac{1}{2})=\frac{\frac{2}{2+\sqrt{5}}}{1+\frac{1}{(2+\sqrt{5})^2}}=\\ \frac{8\sqrt{5}+18}{18\sqrt{5}+40}\\[/latex] [latex]cos(arctg\frac{1}{2})=\frac{1-tg^2\frac{arctg\frac{1}{2}}{2}}{1+tg^2\frac{arctg\frac{1}{2}}{2}}=\\ \frac{1-(\frac{2}{2+\sqrt{5}})^2}{1+(\frac{2}{2+\sqrt{5}})^2}= \frac{4\sqrt{5}+5}{4\sqrt{5}+13} [/latex] [latex]sin(arctg\frac{1}{2}-arctg(-\sqrt{3}))=\\ \frac{8\sqrt{5}+18}{18\sqrt{5}+40}}*cos\frac{\pi}{3}+\frac{4\sqrt{5}+5}{4\sqrt{5}+13} *sin\frac{\pi}{3}=\\ \frac{4\sqrt{5}+9}{18\sqrt{5}+40}+\frac{4\sqrt{5}+5}{4\sqrt{5}+13}*\frac{\sqrt{3}}{2} [/latex]