Question

cos( arctg 1/2- arctg 2)

Answer 1

$cos(arctgfrac{1}{2}-arctg2)=\ cos(arctgfrac{1}{2})cos(arctg2)+sin(arctgfrac{1}{2})sin(arctg2)=\ =|tg^2x+1=frac{1}{cos^2x}==>cos x=frac{1}{sqrt{1+tg^2x}};\ sin^2x=1-cos^2x=1-frac{1}{tg^2x+1}=frac{tg^2x+1-1}{tg^2x+1}=frac{tg^2x}{tg^2x+1};\ sin x=frac{tgx}{sqrt{tg^2x+1}}|$
$=frac{1}{sqrt{1+tg^2(arctgfrac{1}{2})}}cdotfrac{1}{sqrt{1+tg^2(arctg2)}}+frac{tg(arctgfrac{1}{2})}{sqrt{1+tg^2(arctgfrac{1}{2})}}cdotfrac{tg(arctg2)}{sqrt{1+tg^2(arctg2)}}=\ =frac{1}{sqrt{1+(frac{1}{2})^2}}cdotfrac{1}{sqrt{1+2^2}}+frac{frac{1}{2}}{sqrt{1+(frac{1}{2})^2}}cdotfrac{2}{sqrt{1+2^2}}=\ =frac{1}{sqrt{(1+frac{1}{4})(1+4)}}+frac{1}{sqrt{(1+frac{1}{4})(1+4)}}=\ =frac{2}{sqrt{(1+frac{1}{4})(1+4)}}=frac{2}{frac{1}{2}sqrt{(1+4)(1+4)}}=$
$=frac{4}{sqrt{(1+4)^2}}=frac{4}{1+4}=frac{4}{5}$