Question

Решите пожалуйста 4 номер

Answer 1

$1)Cos\alpha =\frac{4}{5}\\\\270^{0}<\alpha<360^{0}\Rightarrow Sin\alpha<0\\\\Sin\alpha=-\sqrt{1-Cos^{2}\alpha}=-\sqrt{1-(\frac{4}{5})^{2} }=-\sqrt{1-\frac{16}{25}}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}$

$2)Sin\alpha =-\frac{3}{5}\\\\180^{0}<\alpha <270^{0}\Rightarrow Cos\alpha<0\\\\Cos\alpha=-\sqrt{1-Sin^{2}\alpha}=-\sqrt{1-(-\frac{3}{5})^{2}}=-\sqrt{1-\frac{9}{25}}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}$