Question

Cos a=-4/5,п/2<a<п.Найти sin a,tg a,ctg a..

Answer 1

$\cos\alpha=-\frac45\\\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac9{25}}=\pm\frac35\\\frac\pi2\ \textless \ \alpha\ \textless \ \pi\Rightarrow\sin\alpha=+\frac35\\tg\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac35:\left(-\frac45\right)=-\frac35\cdot\frac54=-\frac34\\ctg\alpha=\frac1{tg\alpha}=-\frac43=-1\frac13$