Oblicz wartości pozostałych funkcji trygonometrycznych kąta ostrego , jeśli wiadomo że:
[tex]e)\ \sin\alpha=\dfrac{1}{5}\\\\f)\ \cos\alpha=\dfrac{\sqrt2}{6}\\\\h)\ \text{tg}\alpha=\sqrt5[/tex]
[tex]e)\\\cos\alpha=\dfrac{2\sqrt6}{5},\ \text{tg}\alpha=\dfrac{\sqrt6}{12},\ \text{ctg}\alpha=2\sqrt6[/tex]
[tex]f)\\\sin\alpha=\dfrac{\sqrt{34}}{6},\ \text{tg}\alpha=\sqrt{17},\ \text{ctg}\alpha=\dfrac{\sqrt{17}}{17}[/tex]
[tex]h)\\\sin\alpha=\dfrac{\sqrt{30}}{6},\ \cos\alpha=\dfrac{\sqrt6}{6},\ \text{ctg}\alpha=\dfrac{\sqrt5}{5}[/tex]
Jeko, że kąt α jest kątem ostrym, to zadanie możemy rozwiązać na dwa sposoby:
Przykłady e) i f) wykonamy pierwszym sposobem, a h) drugim.
UWAGA: Jako, że kąt α jest kątem ostrym, to wartości wszystkich funkcji trygonometrycznych są dodatnie.
[tex]e)\ \sin\alpha=\dfrac{1}{5}\\\\\left(\dfrac{1}{5}\right)^2+\cos^2\alpha=1\\\\\dfrac{1}{25}+\cos^2\alpha=\dfrac{25}{25}\qquad|-\dfrac{1}{25}\\\\\cos^2\alpha=\dfrac{24}{25}\to\cos\alpha=\pm\sqrt{\dfrac{24}{25}}\\\\\cos\alpha=\pm\dfrac{\sqrt{4\cdot6}}{\sqrt{25}}\\\\\cos\alpha=\pm\dfrac{2\sqrt6}{5}[/tex]
Wartość ujemną odrzucamy.
[tex]\huge\boxed{\cos\alpha=\dfrac{2\sqrt6}{5}}[/tex]
[tex]\text{tg}\alpha=\dfrac{\frac{1}{5}}{\frac{2\sqrt6}{5}}=\dfrac{1}{5}\cdot\dfrac{5}{2\sqrt6}=\dfrac{1}{2\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}=\dfrac{\sqrt6}{2\cdot6}\\\\\huge\boxed{\text{tg}\alpha=\dfrac{\sqrt6}{12}}[/tex]
[tex]\text{ctg}\alpha=\dfrac{12}{\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}=\dfrac{12\sqrt6}{6}\\\\\huge\boxed{\text{ctg}\alpha=2\sqrt6}[/tex]
[tex]f)\ \cos\alpha=\dfrac{\sqrt2}{6}\\\\\sin^2\alpha+\left(\dfrac{\sqrt2}{6}\right)^2=1\\\\\sin^2\alpha+\dfrac{2}{36}=\dfrac{36}{36}\qquad|-\dfrac{2}{36}\\\\\sin^2\alpha=\dfrac{34}{36}\to\sin\alpha=\pm\sqrt{\dfrac{34}{36}}\\\\\sin\alpha=\pm\dfrac{\sqrt{34}}{\sqrt{36}}\\\\\sin\alpha=\pm\dfrac{\sqrt{34}}{6}[/tex]
Wartość ujemną usuwamy.
[tex]\huge\boxed{\sin\alpha=\dfrac{\sqrt{34}}{6}}[/tex]
[tex]\text{tg}\alpha=\dfrac{\frac{\sqrt{34}}{6}}{\frac{\sqrt2}{6}}=\dfrac{\sqrt{34}}{6}\cdot\dfrac{6}{\sqrt2}\\\\\huge\boxed{\text{tg}\alpha=\sqrt{17}}[/tex]
[tex]\text{ctg}\alpha=\dfrac{1}{\sqrt{17}}\cdot\dfrac{\sqrt{17}}{\sqrt{17}}\\\\\huge\boxed{\text{ctg}\alpha=\dfrac{\sqrt{17}}{17}}[/tex]
[tex]h)\ \text{tg}\alpha=\sqrt5=\dfrac{\sqrt5}{1}\to a=\sqrt5,\ b=1\\\\c^2=a^2+b^2\\\\c^2=(\sqrt5)^2+1^2\\\\c^2=5+1\\\\c^2=6\to c=\sqrt6\\\\\sin\alpha=\dfrac{\sqrt5}{\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}\\\\\huge\boxed{\sin\alpha=\dfrac{\sqrt{30}}{6}}\\\\\cos\alpha=\dfrac{1}{\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}\\\\\boxed{\cos\alpha=\dfrac{\sqrt6}{6}}[/tex]
[tex]\text{ctg}\alpha=\dfrac{1}{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}\\\\\huge\boxed{\text{ctg}\alpha=\dfrac{\sqrt5}{5}}[/tex]
