[tex]Wiemy,~~ze~~\alpha ~~jest~~katem~~ostrym~~oraz~~ cos\alpha =\frac{4}{7} \\\\obliczam ~~teraz~~sin\alpha ~~z~~jedynki~~trygonometrycznej:\\\\sin^{2} \alpha +cos^{2} \alpha =1~~\land~~cos\alpha =\frac{4}{7}\\\\~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow\\\\sin^{2} \alpha +(\frac{4}{7} )^{2} =1\\\\sin^{2} \alpha + \frac{16}{49} =1\\\\sin^{2} \alpha = 1 - \frac{16}{49} \\\\sin^{2} \alpha=\frac{33}{49} ~~\land ~~\alpha ~~jest~~katem~~ostrym\\\\sin\alpha =\sqrt{\frac{33}{49} } \\\\[/tex]
[tex]sin\alpha =\frac{\sqrt{33} }{7} \\\\ctg\alpha =\frac{cos\alpha }{sin\alpha } \\\\zad.a\\\\2sin^{2} \alpha - 1 = ~~?~~~\land~~ sin^{2} \alpha =\frac{33}{49} \\\\2\cdot \frac{33}{49} -1 = \frac{66}{49} -\frac{49}{49} =\frac{17}{49} \\\\zad.b\\\\sin^{2} \alpha \cdot ctg^{2} \alpha =sin^{2} \alpha \cdot \frac{cos^{2} \alpha }{sin^{2} \alpha } = cos^{2} \alpha =~~?~~ \land ~~cos\alpha =\frac{4}{7} \\\\(\frac{4}{7} )^{2} =\frac{4}{7} \cdot \frac{4}{7} =\frac{16}{49}[/tex]
[tex]Odp:Gdy~~ cos\alpha =\frac{4}{7}, ~~\alpha ~~jest~~katem~~ostrym~~to: ~~zad.a ~~ 2sin^{2} \alpha -1=\frac{17}{49} , ~~zad.b~~sin^{2} \alpha \cdot ctg^{2} \alpha = \frac{16}{49}[/tex]
