Question

если tgA=1/2 то найдите

Answer 1

$sin(2alpha +frac{pi}{4} )=sin2alpha cosfrac{pi}{4}+cos2alpha sinfrac{pi}{4}=sin2alpha cdotfrac{sqrt{2}}{2}+cos2alpha cdotfrac{sqrt{2}}{2}= \ \ =frac{sqrt{2}}{2}cdot(2sinalpha cosalpha +cos^2alpha -sin^2alpha )=\\=frac{sqrt{2}}{2}cdotcos^2alphacdot(frac{2sinalpha cosalpha}{cos^2alpha}+1 -frac{sin^2alpha }{cos^2alpha})=$

$=frac{sqrt{2}}{2}cdotfrac{1}{frac{1}{cos^2alpha}}cdot(frac{2sinalpha }{cosalpha}+1 -tg^2alpha)= frac{sqrt{2}}{2}cdotfrac{1}{1+tg^2alpha}cdot(2tgalpha+1 -tg^2alpha)= \ \ = frac{sqrt{2}}{2}cdotfrac{1}{1+(frac{1}{2})^2}cdot(2cdotfrac{1}{2}+1- (frac{1}{2})^2)= frac{sqrt{2}}{2}cdotfrac{1}{1+frac{1}{4}}cdot(1+1- frac{1}{4})= \ \ = frac{sqrt{2}}{2}cdotfrac{1}{frac{5}{4}}cdot(2- frac{1}{4})=frac{sqrt{2}}{2}cdotfrac{4}{5}cdotfrac{7}{4}= frac{7sqrt{2}}{10}$