1.
[tex]a = 2\\b = \sqrt{5}[/tex]
Z tw. Pitagorasa obliczam przeciwprostokątną c:
[tex]a^{2}+b^{2} = c^{2}\\\\2^{2} = (\sqrt{5})^{2} = c^{2}\\\\4+5 = c^{2}\\\\c^{2} = 9\\\\c = \sqrt{9}\\\\\underline{c = 3}\\\\sin\alpha = \frac{a}{c} =\boxed{ \frac{2}{3}}\\\\cos\alpha = \frac{b}{c} =\boxed{ \frac{\sqrt{5}}{3}}\\\\tg\alpha = \frac{a}{b} = \frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} = \boxed{\frac{2\sqrt{5}}{5}}[/tex]
2.
Korzystamy z wzorów redukcyjnych:
[tex]sin\alpha = sin(90-\alpha) = cos \alpha[/tex]
oraz z jedynki trygonometrycznej:
[tex]sin^{2}\alpha + cos^{2}\alpha = 1[/tex]
[tex]a) \ sin^{2}15^{o} + sin^{2}75^{o} = sin^{2}15^{o} + sin^{2}(90^{o}-15^{o}) = sin^{2}15^{o}+cos^{2}15^{o} = 1[/tex]
[tex]b) \ (sin62^{o}-cos28^{o})\cdot tg50^{o} = [sin(90^{o}-28^{o})-cos28^{o}]\cdot tg50^{o} =\\\\= (cos28^{o}-cos28^{o})\cdot tg50^{o} =0\cdot tg50^{o} = 0[/tex]
