d)
[tex]cos\alpha=\frac{\sqrt{2}}{2} \ i \ \alpha \in (0^o; \ 180^o) \\\\ sin^2\alpha+ cos^2\alpha=1 \\ sin^2\alpha+ (\frac{\sqrt{2}}{2})^2=1 \\ sin^2\alpha+\frac{2}{4}=1 \\ sin^2\alpha+\frac{1}{2}= 1 \\ sin^2\alpha = 1- \frac{1}{2} \\ sin^2\alpha = \frac{1}{2} \\ sin\alpha = \sqrt{\frac{1}{2}} \in (0^o; \ 180^o) \ lub \ sin\alpha=- \sqrt{\frac{1}{2}} \notin (0^o; \ 180^o) \\ Zatem: \\ sin\alpha=\sqrt{\frac{1}{2}}[/tex]
[tex]sin\alpha= \frac{1}{\sqrt{2}} \\ sin\alpha=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ sin\alpha=\frac{\sqrt{2}}{2} \\\\ tg\alpha = \frac{sin\alpha}{cos\alpha} \\ tg\alpha = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \\ tg\alpha = 1 \\\\ ctg\alpha = \frac{1}{tg\alpha} \\ ctg\alpha = \frac{1}{1} \\ ctg\alpha = 1 \\\\ cos\alpha=\frac{\sqrt{2}}{2}, \ sin\alpha=\frac{\sqrt{2}}{2}, \ tg\alpha=1, \ ctg\alpha=1[/tex]
e)
[tex]cos\alpha=\frac{-5}{8} \ i \ \alpha \in (180^o; \ 270^o) \\\\ sin^2\alpha+ cos^2\alpha=1 \\ sin^2\alpha+ (\frac{-5}{8})^2=1 \\ sin^2\alpha+\frac{25}{64}=1 \\ sin^2\alpha = 1- \frac{25}{64} \\ sin^2\alpha = \frac{39}{64} \\ sin\alpha = \sqrt{\frac{39}{64}} \notin (180^o; \ 270^o) \ lub \ sin\alpha=- \sqrt{\frac{39}{64}} \in (180^o; \ 270^o) \\ Zatem: \\ sin\alpha=-\sqrt{\frac{39}{64}} \\ sin\alpha=-\frac{\sqrt{39}}{8}[/tex]
[tex]tg\alpha =\frac{sin\alpha}{cos\alpha} \\ tg \alpha=\frac{-\frac{\sqrt{39}}{8}}{-\frac{5}{8}} \\ tg \alpha=\frac{\frac{\sqrt{39}}{8}}{\frac{5}{8}} \\ tg\alpha=\frac{\sqrt{39}}{8} : \frac{5}{8} \\ tg\alpha=\frac{\sqrt{39}}{\not{8}_1} \cdot \frac{\not{8}^1}{5} \\ tg\alpha = \frac{\sqrt{39}}{5} \\\\ ctg\alpha = \frac{1}{tg\alpha} \\ ctg\alpha = \frac{1}{\frac{\sqrt{39}}{5}} \\ ctg\alpha = 1:\frac{\sqrt{39}}{5} \\ ctg\alpha=1 \cdot \frac{5}{\sqrt{39}} \\ ctg\alpha=\frac{5}{\sqrt{39}}[/tex]
[tex]ctg\alpha=\frac{5}{\sqrt{39}} \cdot \frac{\sqrt{39}}{\sqrt{39}} \\ ctg\alpha=\frac{5\sqrt{39}}{5} \\\\ cos\alpha =- \frac{5}{8}, \ sin\alpha =-\frac{\sqrt{39}}{8}, \ tg\alpha = \frac{\sqrt{39}}{5}, \ ctg\alpha = \frac{5\sqrt{39}}{5}[/tex]
