2 задачки

[Гост]

Добър ден! Може ли някой да ми помогне с тези 2 задачи, опитвам се да помогна на сина си.

[S.B.]

[tex]sin\alpha = - \frac{3}{5} \Rightarrow cos\alpha = \pm\sqrt{1 - sin^{2}\alpha} = \pm \sqrt{\frac{16}{25}}[/tex]
$\alpha \in (180^\circ ; 270^\circ) \Rightarrow cos\alpha = -\frac{4}{5}$
$sin\displaystyle\frac{\alpha}{2} = \pm\sqrt{\displaystyle\frac{1 - cos\alpha}{2}} = \sqrt{\displaystyle\frac{1 + \displaystyle\frac{4}{5}}{2}} = \pm \sqrt{\displaystyle\frac{9}{10}}$
$\alpha \in (180^\circ ; 270^\circ) \Rightarrow \frac{\alpha}{2} \in (90^\circ; 135^\circ) \Rightarrow sin\frac{\alpha}{2} = \frac{3\sqrt{10}}{10}$
$cos\displaystyle\frac{\alpha}{2} = \pm\sqrt{\displaystyle\frac{1 + cos\alpha}{2}} = \pm \sqrt{\displaystyle\frac{1 - \displaystyle\frac{4}{5}}{2}} = \pm\sqrt{\displaystyle\frac{1}{10}}, \displaystyle \frac{\alpha}{2} \in (90^\circ , 135^\circ) \Rightarrow cos\displaystyle\frac{\alpha}{2} = - \displaystyle\frac{\sqrt{10}}{10}$
$tg\displaystyle\frac{\alpha}{2} = \displaystyle\frac{sin\displaystyle\frac{\alpha}{2}}{cos\displaystyle\frac{\alpha}{2}} = - 3, cotg \displaystyle\frac{\alpha}{2} = \displaystyle\frac{cos\displaystyle\frac{\displaystyle\alpha}{2}}{sin\displaystyle\frac{\displaystyle\alpha}{2}} = - \displaystyle\frac{1}{3}$

[S.B.]

Втора задача:
Щом като [tex]\alpha \in (270^\circ ; 360^\circ)[/tex] то $cos\alpha > 0 , sin\alpha < 0 \Rightarrow$ не е възможно $ cos\alpha = - \frac{1}{3}$
Условието не е коректно зададено.