Question

cos 12°•sin 102° -sin 12°•cos 102°=?
sin100°+cos100°- sin80°+cos80°=?
Cu explicatii plss

Answer 1

[tex]a)sin(a-b)=sina \cdot cos b-sin b\cdot cosa\\ sin102\cdot cos12-sin12 cos102=sin(102-12)=sin100\\ b) sin(180-a)=sina; cos(180-a)=-cosa\\ sin100=sin(180-100)=sin80\\ cos100=-cos(180-100)=-cos80\\ sin 100+cos100-sin80+cos 80=sin80-cos80-sin80+cos80=0[/tex]

Answer 2

1)

Vom folosi formula:

sin(a - b) = sinacosb - sinbcosa

cos12°sin102° - sin12°cos102° = sin 102° cos12° - sin12°cos102° =

= sin(102° - 12°) = sin90° = 1

2)

sin100° =sin(180°-100°) = sin80°

cos100° = - cos(180°-100°) =- cos80°

Expresia din enunț devine:

sin80° - cos80° - sin80° + cos80° = 0