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)

$1-\frac{sin2\alpha cos\alpha }{2sin\alpha } =1-cos\alpha cos\alpha =1-cos^2\alpha =sin^2\alpha$

2)

$\frac{sin2\alpha +sin\alpha }{1+cos2\alpha +cos\alpha } =\frac{2sin\alpha cos\alpha +sin\alpha }{1+cos^2 \alpha -sin^2 \alpha +cos \alpha } =\frac{sin \alpha (2cos \alpha +1)}{2cos^2\alpha +cos a} =\frac{sin \alpha (2cos \alpha +1)}{cos a(2cos a+1)} =\\\\=\frac{sin\alpha }{cos\alpha } =tg\alpha$

3)

$\frac{cos2\alpha }{cos\alpha +sin\alpha } -cos\alpha =\frac{cos^2\alpha -sin^2\alpha }{cos\alpha +sin\alpha} -cos\alpha =\frac{(cos\alpha +sin\alpha)(cos\alpha -sin\alpha)}{cos\alpha +sin\alpha} -cos\alpha =\\=cos\alpha -sin\alpha -cos\alpha =-sin\alpha$

4)

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