Question

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

Answer 1

$sin(arctgfrac{1}{2}-arcctg(-sqrt{3}))=\ sin(arctgfrac{1}{2}+frac{pi}{3})=\ sin(arctgfrac{1}{2})cosfrac{pi}{3}+cos(arctgfrac{1}{2})sinfrac{pi}{3}=$ 
 выразим синус через тангенс 
$sin(arctgfrac{1}{2})cosfrac{pi}{3}+cos(arctgfrac{1}{2})sinfrac{pi}{3}\\ 1)sin(arctgfrac{1}{2})=frac{2tgfrac{arctgfrac{1}{2}}{2}}{1+tg^2frac{arctgfrac{1}{2}}{2}}\ tgfrac{arctgfrac{1}{2}}{2}=frac{tg(arctgfrac{1}{2})}{1+sqrt{1+tg(arctgfrac{1}{2})^2}}=frac{frac{1}{2}}{1+sqrt{1+frac{1}{4}}}=frac{1}{2+sqrt{5}}\ sin(arctgfrac{1}{2})=frac{frac{2}{2+sqrt{5}}}{1+frac{1}{(2+sqrt{5})^2}}=\ frac{8sqrt{5}+18}{18sqrt{5}+40}$
$cos(arctgfrac{1}{2})=frac{1-tg^2frac{arctgfrac{1}{2}}{2}}{1+tg^2frac{arctgfrac{1}{2}}{2}}=\ frac{1-(frac{2}{2+sqrt{5}})^2}{1+(frac{2}{2+sqrt{5}})^2}= frac{4sqrt{5}+5}{4sqrt{5}+13}$

$sin(arctgfrac{1}{2}-arctg(-sqrt{3}))=\ frac{8sqrt{5}+18}{18sqrt{5}+40}}*cosfrac{pi}{3}+frac{4sqrt{5}+5}{4sqrt{5}+13} *sinfrac{pi}{3}=\ frac{4sqrt{5}+9}{18sqrt{5}+40}+frac{4sqrt{5}+5}{4sqrt{5}+13}*frac{sqrt{3}}{2}$