Mamy dane, że α∈⟨90°, 180°⟩ - to jest druga ćwiartka. W drugiej ćwiartce wszystkie funkcje oprócz sinusa przyjmują wartości ujemne.
Będziemy korzystać z tożsamości trygonometrycznych:
[tex]\sin^2x+\cos^2x=1\\\\\text{tg}x=\dfrac{\sin x}{\cos x}[/tex]
[tex]a)\\\sin\alpha=\dfrac{2}{\sqrt6}\\\\\left(\dfrac{2}{\sqrt6}\right)^2+\cos^2\alpha=1\\\\\dfrac{4}{6}+\cos^2\alpha=1\qquad|-\left(\dfrac{4}{6}=\dfrac{2}{3}\right)\\\\\cos^2\alpha=\dfrac{1}{3}\Rightarrow\cos\alpha=\pm\sqrt\dfrac{1}{3}}\\\\\cos\alpha=\pm\dfrac{\sqrt3}{3}\\\\\huge\boxed{\cos\alpha=-\dfrac{\sqrt3}{3}}[/tex]
[tex]\text{tg}\alpha=\dfrac{\frac{2}{\sqrt6}}{-\frac{\sqrt3}{3}}=-\dfrac{2}{\sqrt6}\cdot\dfrac{3}{\sqrt3}=-\dfrac{6}{\sqrt{18}}=-\dfrac{6}{\sqrt{9\cdot2}}=-\dfrac{6}{3\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}=-\dfrac{6\sqrt2}{6}\\\\\huge\boxed{\text{tg}\alpha=-\sqrt2}[/tex]
[tex]b)\ \cos\alpha=-\dfrac{1}{4}\\\\\sin^2\alpha+\left(-\dfrac{1}{4}\right)^2=1\\\\\sin^2\alpha+\dfrac{1}{16}=1\qquad|-\dfrac{1}{16}\\\\\sin^2\alpha=\dfrac{15}{16}\Rightarrow\sin\alpha=\pm\sqrt{\dfrac{15}{16}}\\\\\sin\alpha=\pm\dfrac{\sqrt{15}}{4}\\\\\huge\boxed{\sin\alpha=\dfrac{\sqrt{15}}{4}}[/tex]
[tex]\text{tg}\alpha=\dfrac{\frac{\sqrt{15}}{4}}{-\frac{1}{4}}=-\dfrac{\sqrt{15}}{4}\cdot\dfrac{4}{1}=-\sqrt{15}\\\\\huge\boxed{\text{tg}\alpha=-\sqrt{15}}[/tex]
[tex]c)\ \text{tg}\alpha=-\dfrac{6}{5}\\\\\left\{\begin{array}{ccc}\dfrac{\sin\alpha}{\cos\alpha}=-\dfrac{6}{5}\\\\\sin^2\alpha+\cos^2\alpha=1\end{array}\right[/tex]
Z pierwszego równania mamy:
[tex]\sin\alpha=-\dfrac{6}{5}\cos\alpha[/tex]
Podstawiamy do równania drugiego:
[tex]\left(-\dfrac{6}{5}\cos\alpha\right)^2+\cos^2\alpha=1\\\\\dfrac{36}{25}\cos^2\alpha+\cos^2\alpha=1\\\\\dfrac{36}{25}\cos^2\alpha+\dfrac{25}{25}\cos^2\alpha=1\\\\\dfrac{61}{25}\cos^2\alpha=1\qquad|\cdot\dfrac{25}{61}\\\\\cos^2\alpha=\dfrac{25}{61}\Rightarrow\cos\alpha=\pm\sqrt{\dfrac{25}{61}}\\\\\cos\alpha=\pm\dfrac{5}{\sqrt{61}}\\\\\huge\boxed{\cos\alpha=-\dfrac{5\sqrt{61}}{61}}[/tex]
Podstawiamy do równania pierwszego
[tex]\sin\alpha=-\dfrac{6}{5}\cdot\left(-\dfrac{5\sqrt{61}}{61}\right)\\\\\huge\boxed{\sin\alpha=\dfrac{6\sqrt{61}}{61}}[/tex]