Question

Помогите решить,даю 70 баллов)

Answer 1

Ответ:

1)sina^2

2)sin2a+sina/1+2cosa^2-1+cosa

3)-sina

4)cos2°

Объяснение:

1)...

1-sin2acosa/2sina

1-cosacosa

1-cosa^2

sina^2

4)...

2sin30°cos(-2°)

2•1/2•cos2°

cos2°

Answer 2

1)

$1-frac{sin2alpha cosalpha }{2sinalpha } =1-cosalpha cosalpha =1-cos^2alpha =sin^2alpha$

2)

$frac{sin2alpha +sinalpha }{1+cos2alpha +cosalpha } =frac{2sinalpha cosalpha +sinalpha }{1+cos^2 alpha -sin^2 alpha +cos alpha } =frac{sin alpha (2cos alpha +1)}{2cos^2alpha +cos a} =frac{sin alpha (2cos alpha +1)}{cos a(2cos a+1)} =\\=frac{sinalpha }{cosalpha } =tgalpha$

3)

$frac{cos2alpha }{cosalpha +sinalpha } -cosalpha =frac{cos^2alpha -sin^2alpha }{cosalpha +sinalpha} -cosalpha =frac{(cosalpha +sinalpha)(cosalpha -sinalpha)}{cosalpha +sinalpha} -cosalpha =\=cosalpha -sinalpha -cosalpha =-sinalpha$

4)

$sin(-28а)+sin(-32а)=2sin30а*cos(-2а)=2*frac{1}{2} *cos2а=cos2а$