Rozwiąże ktoś? (zdjęcie)

Rozwiąże Ktoś Zdjęcie class=

Odpowiedź :

[tex]zad.2\\ \\ sin\alpha =\dfrac{2}{5} ~~\land ~~\alpha \in (90^{0} ,180^{0} )\\ \\ sin^{2} \alpha +cos^{2} \alpha =1~~\land~~sin\alpha =\dfrac{2}{5} \\ \\ cos^{2} \alpha +(\frac{2}{5} )^{2} =1\\ \\ cos^{2} \alpha=1-\dfrac{4}{25} \\ \\ cos^{2} \alpha=\dfrac{21}{25} ~~\land~~\alpha \in (90^{0} ,180^{0} )~~\Rightarrow ~~cos\alpha =-\dfrac{\sqrt{21} }{5} \\ \\ tg\alpha =\dfrac{sin\alpha }{cos\alpha } ~~\land~~cos\alpha =-\dfrac{\sqrt{21} }{5} ~~\land ~~sin\alpha =\dfrac{2}{5}\\ \\ \\ [/tex]

[tex]tg\alpha =\dfrac{\dfrac{2}{5} }{-\dfrac{\sqrt{21} }{5} } =-\dfrac{2}{\sqrt{21} } \\ \\ tg\alpha =-\dfrac{2}{\sqrt{21} } \cdot \dfrac{\sqrt{21} }{\sqrt{21} }\\ \\\\ tg\alpha =-\dfrac{2\sqrt{21} }{21} [/tex]

[tex]Pamietaj:~~II~~cwiartka~~tylko~~sinus~~jest~~dodatni.[/tex]

[tex]zad.1\\ \\ a_{n} =a_{1} \cdot q^{n-1} ~~~~n-ty ~~wyraz~~ciagu~~geometrycznego\\ \\ a_{2} =a_{1} \cdot q~~\land ~~a_{2} =9 ~~\Rightarrow ~~a_{1} \cdot q=9\\ \\ a_{4} =a_{1} \cdot q^{3} ~~\land ~~a_{4} =1 ~~\Rightarrow ~~a_{1} \cdot q^{3} =1\\ \\ a_{1} \cdot q^{3} =1\\ \\ a_{1} \cdot q\cdot q^{2} =1~~\land ~~a_{1} \cdot q=9~~\Rightarrow ~~9\cdot q^{2} =1\\ \\ 9\cdot q^{2} =1~~\mid \div 9\\ \\ q^{2}=\dfrac{1}{9}\\ \\ q=\dfrac{1}{3} ~~\lor~~q=-\dfrac{1}{3} \\ \\ \\ [/tex]

[tex]a_{1} \cdot q=9~~\Rightarrow ~~a_{1} =\dfrac{9}{q} \\ \\ (~~q=\dfrac{1}{3} ~~\land ~~a_{1} =27~~)~~\lor ~~(~~q=-\dfrac{1}{3} ~~\land ~~a_{1} =-27~~)[/tex]