Question

sin 160 градус cos 110 градус + sin 250 градус cos 340 градус + tg 110 градус tg 340 градус​

Answer 1

$sin160^o*cos110^o+sin250^o*cos340^o+tg110^o*tg340^o=\\\\=sin(180^o-20^o)*cos(90^o+20^o)+sin(270^o-20^o)*cos(360^o-20^o)+\\\\+tg(90^o+20^o)*tg(360^o-20^o)=sin20^o*(-sin20^o)+(-cos20^o)*cos20^o+\\\\+(-ctg20^o)*(-tg20^o)=-sin^220^o-cos^220^o+ctg20^o*tg20^o=\\\\=-(sin^220^o+cos^220^o)+ctg20^o*tg20^o=-1+1=0\\\\Formyla\::\\\\\boxed{\:\:sin^2\alpha+cos^2\alpha=1\;\;}\\\\\boxed{\;\;ctg\alpha*tg\alpha=1\;\;}\\\\Otvet\::\;\;\boldsymbol{0}$