Question

помогите пожалуйста, 40 баллов

Answer 1

$frac{sin4alpha}{cos3alpha-cosalpha}=frac{2sin2alphacos2alpha}{cos(2alpha+alpha)-cosalpha}=frac{4sinalphacosalphacos2alpha}{cos2alphacosalpha-sin2alphasinalpha-cosalpha}=$
$=frac{4sinalphacosalphacos2alpha}{cos2alphacosalpha-2cosalphasin^2alpha-cosalpha}=4frac{cosalpha}{cosalpha}cdotfrac{sinalphacos2alpha}{cos2alpha-2sin^2alpha-1}=\ =4frac{sinalphacos2alpha}{1-2sin^2alpha-2sin^2alpha-1}=frac{4}{-4}cdotfrac{sinalphacos2alpha}{sin^2alpha}=-frac{cos2alpha}{sinalpha}=\ =-frac{cos^2alpha-sin^2alpha}{sinalpha}=sinalpha-cosalphacdot ctgalpha$

Answer 2

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