Question

Тригонометрия. Помогите, пожалуйста!

Answer 1

$1.;frac12sin4alpha(ctgalpha-tgalpha)=sin2alphacos2alphaleft(frac{cosalpha}{sinalpha}-frac{sinalpha}{cosalpha}right)=\=sin2alphacos2alphacdotfrac{cos^2alpha-sin^2alpha}{sinalphacosalpha}=sin2alphacos2alphacdotfrac{cos2alpha}{frac12sin2alpha}=\=sin2alphacos2alphacdotfrac{2cos2alpha}{sin2alpha}=2cos^22alpha=(2cos^22alpha-1)+1=cos4alpha+1$
$2.;a);cos(alpha-beta)-cos(alpha+beta)=(cosalphacosbeta+sinalphasinbeta)-\-( (cosalphacosbeta-sinalphasinbeta))= cosalphacosbeta+sinalphasinbeta-\-cosalphacosbeta+sinalphasinbeta=0$
$b);frac{sin(-alpha)+cos(pi+alpha)}{2cosleft(fracpi2- alpha right)cos(-alpha)+1}=frac{-sinalpha -cosalpha }{2sinalpha cosalpha +1}=frac{-(sinalpha+cosalpha)}{2sinalpha cosalpha +sin^2alpha +cos^2alpha }=\=frac{-(sinalpha+cosalpha)}{( sinalpha +cosalpha)^2 }=-frac{1}{ sinalpha +cosalpha }\3.;4cosfrac{3alpha}2cosalphasinfracalpha2=4sinfracalpha2cosfrac{3alpha}2cosalpha=4cdotfrac{sin(-alpha)+sin2alpha}2cosalpha=$
$=2cdot({-sinalpha+sin2alpha})cdotcosalpha=2sin2alphacosalpha-2sinalphacosalpha=\=2sin2alphacosalpha-sin2alpha=sin2alpha(2cosalpha-1)$
$4.;frac{cos3alpha+cos2alpha+cosalpha+1}{cosalpha+2cos^2fracalpha2-1}=2cosfrac{3alpha}2cosfracalpha2\frac{cos3alpha+cos2alpha+cosalpha+1}{cosalpha+2cos^2fracalpha2-1}=frac{4cos^3alpha-3cosalpha+2cos^2alpha-1+cosalpha+1}{cosalpha+cosalpha}=\=frac{4cos^3alpha-2cosalpha+2cos^2alpha}{2cosalpha}=frac{2cosalpha(cos^2alpha-1)+2cos^2alpha}{2cosalpha}=$
$=frac{2cosalphacos2alpha+2cos^2alpha}{2cosalpha}=frac{2cosalpha(cos2alpha+cosalpha)}{2cosalpha}=cos2alpha+cosalpha\2cosfrac{3alpha}2cosfrac{alpha}2=2cdotfrac{cosalpha+cos2alpha}2=cosalpha+cos2alpha\{cosalpha+cos2alpha}=cosalpha+cos2alpha$