Odpowiedź:
a)
Wiemy, że
[tex]sin\alpha = \frac{2}{5}[/tex]
[tex]sin^2\alpha +cos^2\alpha =1[/tex]
Zatem:
[tex]cos\alpha = \sqrt{1-sin^2\alpha } = \sqrt{1-\frac{4}{25} } = \frac{\sqrt{21} }{5}[/tex]
[tex]tg\alpha = \frac{sin\alpha }{cos\alpha } = \frac{\frac{2}{5} }{\frac{\sqrt{21} }{5} } = \frac{2}{\sqrt{21} } = \frac{2\sqrt{21}}{21}[/tex]
[tex]ctg\alpha = \frac{1}{tg\alpha } = \frac{\sqrt{21} }{2}[/tex]
b)
[tex]tg \alpha = 2 \\ctg\alpha = \frac{1}{tg\alpha } = \frac{1}{2}[/tex]
[tex]tg\alpha = \frac{sin\alpha }{cos\alpha } \\\frac{sin\alpha }{cos\alpha } = 2\\sin\alpha =2cos\alpha[/tex]
Teraz znowu 1. trygonometryczna.
[tex]sin^2\alpha +cos^2\alpha =1\\(2cos\alpha )^2 + cos^2\alpha =1\\5cos^2\alpha =1\\cos\alpha = \sqrt{\frac{1}{5} } = \frac{\sqrt{5} }{5}[/tex]
[tex]sin\alpha =2cos\alpha \\sin\alpha =2*\frac{\sqrt{5} }{5} = \frac{2\sqrt{5} }{5}[/tex]
Pozdrawiam:)
