Odpowiedź :
Aby obliczyć pozostałe wartości funkcji trygonometrycznych będę korzystała ze wzorów:
[tex]tg\alpha =\frac{1}{ctg\alpha } \\\\ctg\alpha =\frac{1}{tg\alpha } \\\\tg\alpha =\frac{sin\alpha }{cos\alpha } \\\\ctg\alpha =\frac{cos\alpha }{sin\alpha } \\\\sin^{2} \alpha +cos^{2} \alpha =1[/tex]
[tex]tg\alpha =\frac{5}{12} ~~\\\\tg\alpha =\frac{5}{12} ~~\land ~~ctg\alpha =\frac{1}{tg\alpha } ~~\Rightarrow ~~ctg\alpha =\frac{12}{5} \\\\ctg\alpha =2\frac{2}{5} =2,4\\\\tg\alpha =\frac{5}{12} ~~\land ~~tg\alpha =\frac{sin\alpha }{cos\alpha } ~~\Rightarrow ~~\frac{sin\alpha }{cos\alpha }=\frac{5}{12}\\\\sin\alpha =\frac{5}{12}cos\alpha ~~\land ~~sin^{2} \alpha +cos^{2}\alpha =1\\\\ (\frac{5}{12}cos\alpha )^{2} +cos^{2}\alpha =1\\\\\frac{25}{144} cos^{2} \alpha +cos^{2}\alpha =1[/tex]
[tex]\frac{169}{144} cos^{2} \alpha =1~~\mid\div \frac{169}{144}\\\\cos^{2} \alpha =\frac{144}{169}~~\land ~~\alpha -kat~~ostry~~\Rightarrow cos\alpha =\frac{12}{13} \\\\cos\alpha =\frac{12}{13} ~~\land ~~sin\alpha =\frac{5}{12} cos\alpha ~~\Rightarrow ~~sin\alpha =\frac{5}{13}[/tex]
[tex]ctg\alpha =1\frac{7}{8} \\\\ctg\alpha =\frac{15}{8}~~\land ~~tg\alpha =\frac{1}{ctg\alpha } ~~\Rightarrow ~~tg\alpha =\frac{8}{15} \\\\tg\alpha =\frac{sin\alpha }{cos\alpha } ~~\land~~tg\alpha =\frac{8}{15} \Rightarrow ~~\frac{8}{15}=\frac{sin\alpha }{cos\alpha } \\\\sin\alpha =\frac{8}{15} cos\alpha ~~\land~~sin^{2} \alpha +cos^{2} \alpha =1\\\\(\frac{8}{15} cos\alpha)^{2} +cos^{2} \alpha =1\\\\\frac{64}{225} cos^{2} \alpha +cos^{2} \alpha =1\\\\[/tex]
[tex]\frac{289}{225} cos^{2} \alpha =1~~\mid~~\div \frac{289}{225} \\\\cos^{2} \alpha =\frac{225}{289} ~~\land ~~\alpha -kat~~ostry\\\\cos\alpha =\frac{15}{17} \\\\sin\alpha =\frac{8}{15} cos\alpha ~~\land ~~cos\alpha =\frac{15}{17}~~\Rightarrow ~~sin\alpha =\frac{8}{17}[/tex]
[tex]ctg\alpha =2 \\\\ctg\alpha =\frac{2}{1}~~\land ~~tg\alpha =\frac{1}{ctg\alpha } ~~\Rightarrow ~~tg\alpha =\frac{1}{2} \\\\tg\alpha =\frac{sin\alpha }{cos\alpha } ~~\land~~tg\alpha =\frac{1}{2} \Rightarrow ~~\frac{1}{2}=\frac{sin\alpha }{cos\alpha } \\\\sin\alpha =\frac{1}{2} cos\alpha ~~\land~~sin^{2} \alpha +cos^{2} \alpha =1\\\\(\frac{1}{2} cos\alpha)^{2} +cos^{2} \alpha =1\\\\\frac{1}{4} cos^{2} \alpha +cos^{2} \alpha =1\\\\[/tex]
[tex]\frac{5}{4} cos^{2} \alpha =1~~\mid~~\div \frac{5}{4} \\\\cos^{2} \alpha =\frac{4}{5} ~~\land ~~\alpha -kat~~ostry\\\\cos\alpha =\frac{2}{\sqrt{5} } =\frac{2\sqrt{5} }{5} \\\\sin\alpha =\frac{1}{2} cos\alpha ~~\land ~~cos\alpha =\frac{2\sqrt{5} }{5}~~\Rightarrow ~~sin\alpha =\frac{\sqrt{5} }{5}[/tex]
Odpowiedź:
Zadanie wykonam w załączniku
Szczegółowe wyjaśnienie: