Question

Найдите sin2a и cos2a , если sina=-2/3 , П<a<3П/2

Answer 1

 sina=-2/3 ; П<a<3П/2
cosa=-√(1-(-2/3)²=-√5/3
sin2a=2·sina·cosa=2·(-2/3)·(-√5/3)=(4√5)/9
cos2a=cos²a-sin²a=(-√5/3)²-((4√5)/9)²=5/9-80/81=45/81-80/81=35/81