Question

Срочно!!!!! Дам 50 баллов.
Только с нормальным решением

Answer 1

$1);sinalphasin(alpha+beta)+cosalphacos(alpha+beta)=\=sinalpha(sinalphacosbeta+cosalphasinbeta)+cosalpha(cosalphacosbeta-sinalphasinbeta)=\=sin^2alphacosbeta+sinalphacosalphasinbeta+cos^2alphasinbeta-sinalphacosalphasinbeta=\=sin^2alphacosbeta+cos^2alphasinbeta$

$2);frac{cos(alpha+beta)+cos(alpha-beta)}{cos(alpha-beta)-cos(alpha+beta)}=frac{cosalphacosbeta-sinalphasinbeta+cosalphacosbeta+sinalphasinbeta}{cosalphacosbeta+sinalphasinbeta-cosalphacosbeta+sinalphasinbeta}=\\=frac{2cosalphacosbeta}{2sinalphasinbeta}=ctgalphacdot ctgbeta$

$3);frac{sin11^ocos15^o+cos11^osin15^o}{sin18^ocos12^o+cos18^osin12^o}=frac{sin(11^o+15^o)}{sin(18^o+12^o)}=frac{sin26^o}{sin30^o}=2sin26^o$

$4);frac{tg7alpha-tg3alpha}{tg7alphacdot tg3alpha+1}=tg(7alpha-3alpha)=tg4alpha\\5);cos^2alpha-4sin^2fracalpha2cos^2fracalpha2=cos^2alpha-(2sinfracalpha2cosfracalpha2)^2=cos^2alpha-(sinalpha)^2=\=cos^2alpha-sin^2alpha=cos2alpha$

$6);sqrt{1+cos8alpha}=sqrt{1+2cos^24alpha-1}=sqrt{2cos^24alpha}=sqrt2cdotcos4alpha\\7);frac{sinalphacosalpha}{1-2sin^2alpha}=frac{frac12sin2alpha}{cos2alpha}=frac12tg2alpha$