Odpowiedź:
[tex]\huge\boxed{sin\alpha = \frac{3\sqrt{5}}{7} \ \ \ | \ \ \ tg\alpha = \frac{3\sqrt{5}}{2}}[/tex]
[tex]0^{o} < \alpha < 90^{o}\\cos\alpha = \frac{2}{7}\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha = 1-cos^{2}\alpha = 1-(\frac{2}{7})^{2} = 1-\frac{4}{49} = \frac{49}{49}-\frac{4}{49} = \frac{45}{49}\\\\sin\alpha = \sqrt{\frac{45}{49}} = \frac{\sqrt{45}}{\sqrt{49}} = \frac{\sqrt{9\cdot5}}{7} =\boxed{\frac{3\sqrt{5}}{7}}[/tex]
[tex]tg\alpha = \frac{sin\alpha}{cos\alpha} = \frac{\frac{3\sqrt{5}}{7}}{\frac{2}{7}} = \frac{3\sqrt{5}}{7}\cdot\frac{7}{2} = \boxed{\frac{3\sqrt{5}}{2}}[/tex]
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Odpowiedź:
[tex]\huge\boxed{sin\alpha = \frac{3\sqrt{5}}{7} \ \ \ | \ \ \ tg\alpha = \frac{3\sqrt{5}}{2}}[/tex]
Funkcje trygonometryczne
[tex]0^{o} < \alpha < 90^{o}\\cos\alpha = \frac{2}{7}\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha = 1-cos^{2}\alpha = 1-(\frac{2}{7})^{2} = 1-\frac{4}{49} = \frac{49}{49}-\frac{4}{49} = \frac{45}{49}\\\\sin\alpha = \sqrt{\frac{45}{49}} = \frac{\sqrt{45}}{\sqrt{49}} = \frac{\sqrt{9\cdot5}}{7} =\boxed{\frac{3\sqrt{5}}{7}}[/tex]
[tex]tg\alpha = \frac{sin\alpha}{cos\alpha} = \frac{\frac{3\sqrt{5}}{7}}{\frac{2}{7}} = \frac{3\sqrt{5}}{7}\cdot\frac{7}{2} = \boxed{\frac{3\sqrt{5}}{2}}[/tex]