Question

помогите,пожалуйста, с подробным решением

Answer 1

$frac{sin(frac{3pi }{2}+alpha)+sin(2pi+alpha)}{2cos(-alpha)*sin(-alpha)+1}=frac{-cosalpha+sinalpha}{-2cosalpha*sinalpha+1}=frac{sinalpha-cosalpha}{1-2cosalpha*sinalpha}=$$=frac{sinalpha-cosalpha}{cos^2alpha+sin^2alpha-2cosalpha*sinalpha}=frac{sinalpha-cosalpha}{(sinalpha-cosalpha)^2}=frac{1}{sinalpha-cosalpha}$

P.S.

$cos(-alpha)=cosalpha$

$sin(-alpha)=-sinalpha$

$sin(2pi +alpha)=sinalpha$

$sin(frac{3pi }{2}+alpha)=-cosalpha$

$frac{1}{sinalpha-cosalpha}=frac{frac{sqrt{2} }{2} }{frac{sqrt{2} }{2}(sinalpha-cosalpha)} =frac{frac{sqrt{2} }{2} }{frac{sqrt{2} }{2}*sinalpha-frac{sqrt{2} }{2} cosalpha} =frac{frac{sqrt{2} }{2} }{cosfrac{pi }{4}*sinalpha-sinfrac{pi }{4} cosalpha} =$$=frac{frac{sqrt{2}}{2}}{sin(alpha-frac{pi }{4})}=frac{sqrt{2} }{2sin(alpha-frac{pi }{4})}=frac{1}{sqrt{2}sin(alpha-frac{pi }{4})}$