Question

Ребят, помогите решить, срочно нужно:
$frac{sin72^0cos 12^0-sin12^0 cos72^0}{cos18^0cos12^0-sin18^0sin12^0}$

Answer 1

$\sinalphacosbeta-sinbetacosalpha=sin(alpha-beta)\cosalphacosbeta-sinalphasinbeta=cos(alpha+beta)\frac{sin72^ocos12^o-sin12^ocos72^o}{cos18^ocos12^o-sin18^osin18^o}=frac{sin(72^o-12^o)}{cos(18^o+12^o)}=frac{sin60^o}{cos30^o}=frac{frac1{sqrt2}}{frac1{sqrt2}}=1$