Zadanie 10.
Wiedząc, że cos\alpha=1/6 oblicz tg\alpha.
cos\alpha=\frac{1}{6} , tg\alpha=?
z jedynki trygonometrycznej
sin^2\alpha+cos^2\alpha=1
sin^2\alpha+(\frac{1}{6})^2=1
sin^2\alpha+\frac{1}{36}=1
sin^2\alpha=\frac{36}{36}-\frac{1}{36}
sin^2\alpha=\frac{35}{36}
sin\alpha=-\sqrt{\frac{35}{36}} lub sin\alpha=\sqrt{\frac{35}{36}}
sin\alpha=-\frac{\sqrt{35}}{6} lub sin\alpha=\frac{\sqrt{35}}{6}
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tg\alpha=\frac{sin\alpha}{cos\alpha}
tg\alpha=\frac{-\frac{\sqrt{35}}{6}}{\frac{1}{6}}=-\frac{\sqrt{35}}{\not6^1}*\frac{\not6^1}{1}=-\sqrt{35}
lub
tg\alpha=\frac{\frac{\sqrt{35}}{6}}{\frac{1}{6}}=\frac{\sqrt{35}}{\not6^1}*\frac{\not6^1}{1}=\sqrt{35}
Odpowiedź:
tg\alpha=-\sqrt{35} lub tg\alpha=\sqrt{35}