Question

Помогите решить А и Б пожалуйста

Answer 1

$a); frac{1+cos(frac{3pi}{2}+ alpha )+cos(pi +2 alpha )+cos(frac{pi}{2}+3 alpha )}{2sin^2 alpha +sin alpha -1} =\\= frac{1+sin alpha -cos2 alpha -sin3 alpha }{2sin^2 alpha +sin alpha -1} = frac{2sin^2 alpha +sin alpha -(3sin alpha -4sin^3 alpha ) }{2sin^2 alpha +sin alpha -1} =\\= frac{2sin^2 alpha -2sin alpha +4sin^3 alpha }{2sin^2 alpha +sin alpha -1} = frac{2sin alpha (2sin^2 alpha +sin alpha -1)}{2sin^2 alpha +sin alpha -1} =2sin alpha$

$b); ; (frac{1}{cos3 alpha }+frac{1}{cos alpha })cdot (frac{cos2 alpha }{sin alpha }-frac{sin2 alpha }{cos alpha} )cdot (frac{ctg alpha -tg alpha }{1+cos4 alpha })=\\= frac{cos alpha +cos3 alpha }{cos3 alpha cdot cos alpha } cdot frac{cos2 alpha cdot cos alpha -sin2 alpha cdot sin alpha }{sin alpha cdot cos alpha } cdot frac{cos^2 alpha -sin^2 alpha }{sinalpha cdot cos alpha cdot 2cos^22 alpha }=$

$= frac{2cos2 alpha cdot cos alpha }{cos alpha cdot sin alpha } cdot frac{cos3 alpha }{sin alpha cdot cos alpha } cdot frac{cos2 alpha }{sin alpha dot cos alpha cdot 2cos^22 alpha } =\\= frac{1}{(sin alpha cdot cos alpha )^2} =frac{1}{(frac{1}{2}sin2 alpha )^2}=frac{4}{sin^22 alpha }$