$a\text{)}\sin \alpha =-\frac{3}{5}\text{i}{270}^{\circ }<\alpha <{360}^{\circ }$  


Korzystając z jedynki trygonometrycznej wyznaczmy wartość  $\cos \alpha$  :

$\sin {\alpha }^2+\cos {\alpha }^2=1$         

${\left(-\frac{3}{5}\right)}^2+\cos {\alpha }^2=1$  

$\frac{9}{25}+\cos {\alpha }^2=1$  

$\cos {\alpha }^2=\frac{16}{25}$  

$\cos \alpha =\frac{4}{5}$   

ponieważ:

$\alpha \in \left({270}^{\circ };{360}^{\circ }\right)$  


Obliczmy wartość  $\mathrm{tg}\alpha$  :

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }$  

$\mathrm{tg}\alpha =\frac{-\frac{3}{5}}{\frac{4}{5}}=-\frac{3}{{\sout{5}}_1}\cdot \frac{{\sout{5}}^1}{4}=-\frac{3}{4}$    

---

$b\text{)}\cos \alpha =\frac{2}{3}\text{i}-{90}^{\circ }<\alpha <0^{\circ }$  


Korzystając z jedynki trygonometrycznej wyznaczmy wartość  $\sin \alpha$  :

$\sin {\alpha }^2+\cos {\alpha }^2=1$         

$\sin {\alpha }^2+{\left(\frac{2}{3}\right)}^2=1$  

$\sin {\alpha }^2+\frac{4}{9}=1$  

$\sin {\alpha }^2=\frac{5}{9}$  

$\sin \alpha =-\frac{\sqrt{5}}{3}$  

ponieważ:

$\alpha \in \left(-{90}^{\circ };0^{\circ }\right)$  


Obliczmy wartość  $\mathrm{tg}\alpha$  :

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }$  

$\mathrm{tg}\alpha =\frac{-\frac{\sqrt{5}}{3}}{\frac{2}{3}}=-\frac{\sqrt{5}}{{\sout{3}}_1}\cdot \frac{{\sout{3}}^1}{2}=-\frac{\sqrt{5}}{2}$   

---

$c\text{)}\sin \alpha =\frac{1}{4}\text{i}{0}^{\circ }<\alpha <{90}^{\circ }$  


Korzystając z jedynki trygonometrycznej wyznaczmy wartość  $\cos \alpha$  :

$\sin {\alpha }^2+\cos {\alpha }^2=1$         

${\left(\frac{1}{4}\right)}^2+\cos {\alpha }^2=1$  

$\frac{1}{16}+\cos {\alpha }^2=1$  

$\cos {\alpha }^2=\frac{15}{16}$  

$\cos \alpha =\frac{\sqrt{15}}{4}$  

ponieważ:

$\alpha \in \left(0^{\circ };{90}^{\circ }\right)$  


Obliczmy wartość  $\mathrm{tg}\alpha$  :

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }$  

$\mathrm{tg}\alpha =\frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}}=\frac{1}{4}\cdot \frac{4}{\sqrt{15}}=\frac{1}{\sqrt{15}}=\frac{\sqrt{15}}{15}$  

---

$d\text{)}\cos \alpha =\frac{1}{5}\text{i}-{90}^{\circ }<\alpha <0^{\circ }$  


Korzystając z jedynki trygonometrycznej wyznaczmy wartość  $\sin \alpha$  :

$\sin {\alpha }^2+\cos {\alpha }^2=1$         

$\sin {\alpha }^2+{\left(\frac{1}{5}\right)}^2=1$  

$\sin {\alpha }^2+\frac{1}{25}=1$  

$\sin {\alpha }^2=\frac{24}{25}$  

$\sin \alpha =-\frac{2\sqrt{6}}{5}$  

ponieważ:

$\alpha \in \left(-{90}^{\circ };0^{\circ }\right)$  


Obliczmy wartość  $\mathrm{tg}\alpha$  :

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }$  

$\mathrm{tg}\alpha =\frac{\frac{-2\sqrt{6}}{5}}{\frac{1}{5}}=\frac{-2\sqrt{6}}{{\sout{5}}_1}\cdot \frac{{\sout{5}}^1}{1}=-2\sqrt{6}$  

---

$e\text{)}\mathrm{tg}\alpha =3\text{i}-{360}^{\circ }<\alpha <-{270}^{\circ }$  


wiemy, że:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }$  

zatem:

$\{\begin{array}{l}\sin {\alpha }^2+\cos {\alpha }^2=1\\ \frac{\sin \alpha }{\cos \alpha }=3\end{array}$  

$\{\begin{array}{l}\sin {\alpha }^2+\cos {\alpha }^2=1\\ \sin \alpha =3\cos \alpha \end{array}$  

$\{\begin{array}{l}{\left(3\cos \alpha \right)}^2+\cos {\alpha }^2=1\\ \sin \alpha =3\cos \alpha \end{array}$  

$\{\begin{array}{l}9\cos {\alpha }^2+\cos {\alpha }^2=1\\ \sin \alpha =3\cos \alpha \end{array}$  

$\{\begin{array}{l}10\cos {\alpha }^2=1\\ \sin \alpha =3\cos \alpha \end{array}$  

$\{\begin{array}{l}\cos {\alpha }^2=\frac{1}{10}\\ \sin \alpha =3\cos \alpha \end{array}$  

$\{\begin{array}{l}\cos \alpha =\frac{\sqrt{10}}{10}\\ \sin \alpha =3\cos \alpha \end{array}$  

$\{\begin{array}{l}\cos \alpha =\frac{\sqrt{10}}{10}\\ \sin \alpha =\frac{3\sqrt{10}}{10}\end{array}$  

ponieważ:

$\alpha \in \left(-{360}^{\circ };-{270}^{\circ }\right)$  

---

$f\text{)}\mathrm{tg}\alpha =-\frac{1}{2}\text{i}{90}^{\circ }<\alpha <{180}^{\circ }$  


wiemy, że:

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }$  

zatem:

$\{\begin{array}{l}\sin {\alpha }^2+\cos {\alpha }^2=1\\ \frac{\sin \alpha }{\cos \alpha }=-\frac{1}{2}\end{array}$  

$\{\begin{array}{l}\sin {\alpha }^2+\cos {\alpha }^2=1\\ \sin \alpha =-\frac{1}{2}\cos \alpha \end{array}$  

$\{\begin{array}{l}{\left(-\frac{1}{2}\cos \alpha \right)}^2+\cos {\alpha }^2=1\\ \sin \alpha =-\frac{1}{2}\cos \alpha \end{array}$  

$\{\begin{array}{l}\frac{1}{4}\cos {\alpha }^2+\cos {\alpha }^2=1\\ \sin \alpha =-\frac{1}{2}\cos \alpha \end{array}$  

$\{\begin{array}{l}\frac{5}{4}\cos {\alpha }^2=1\\ \sin \alpha =-\frac{1}{2}\cos \alpha \end{array}$  

$\{\begin{array}{l}\cos {\alpha }^2=\frac{4}{5}\\ \sin \alpha =-\frac{1}{2}\cos \alpha \end{array}$  

$\{\begin{array}{l}\cos \alpha =\frac{2}{\sqrt{5}}\\ \sin \alpha =-\frac{1}{2}\cos \alpha \end{array}$  

$\{\begin{array}{l}\cos \alpha =\frac{-2\sqrt{5}}{5}\\ \sin \alpha =-\frac{1}{2}\cdot \frac{-2\sqrt{5}}{5}\end{array}$  

$\{\begin{array}{l}\cos \alpha =\frac{-2\sqrt{5}}{5}\\ \sin \alpha =\frac{\sqrt{5}}{5}\end{array}$  

ponieważ:

$\alpha \in \left({90}^{\circ };{180}^{\circ }\right)$