Szczegółowe wyjaśnienie:
Wiemy:
[tex]\sin^2x+\cos^2x=1\\\\\text{ctg}x=\dfrac{\cos x}{\sin x}=\dfrac{1}{\text{tg}x}\\\\\text{tg}x=\dfrac{\sin x}{\cos x}=\dfrac{1}{\text{ctg}x}[/tex]
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[tex]b)\ \text{ctg}\alpha=\dfrac{8}{15},\ \alpha\in(180^o,\ 270^o)\Rightarrow\ \sin\alpha<0\ \wedge\ \cos\alpha<0\\\\\huge\boxed{\text{tg}\alpha=\dfrac{15}{8}}[/tex]
[tex]\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{8}{15}\Rightarrow\sin\alpha=\dfrac{8}{15}\cos\alpha[/tex]
podstawiamy do jedynki trygonometrycznej:
[tex]\left(\dfrac{8}{15}\cos\alpha\right)^2+\cos^2\alpha=1\\\\\dfrac{64}{225}\cos^2\alpha+\cos^2\alpha=1\\\\\dfrac{64}{225}\cos^2\alpha+\dfrac{225}{225}\cos^2\alpha=1\\\\\dfrac{289}{225}\cos^2\alpha=1\qquad|\cdot\dfrac{225}{289}\\\\\cos^2\alpha=\dfrac{225}{289}\to\cos\alpha=\pm\sqrt{\dfrac{225}{289}}\\\\\cos\alpha=\pm\dfrac{15}{17}\\\\\huge\boxed{\cos\alpha=-\dfrac{15}{17}}[/tex]
[tex]\sin\alpha=\dfrac{8}{15}\cdot\left(-\dfrac{15}{17}\right)\\\\\huge\boxed{\sin\alpha=-\dfrac{8}{17}}[/tex]
[tex]c)\ \cos\alpha=\dfrac{2}{3},\ \alpha\in(270^o,\ 360^o)\Rightarrow\sin\alpha<0\ \wedge\ \text{tg}\alpha<0\ \wedge\ \text{ctg}\alpha<0[/tex]
podstawiamy do jedynki trygonometrycznej:
[tex]\sin^2\alpha+\left(\dfrac{2}{3}\right)^2=1\\\\\sin^2\alpha+\dfrac{4}{9}=1\qquad|-\dfrac{4}{9}\\\\\sin^2\alpha=\dfrac{5}{9}\to\sin\alpha=\pm\sqrt{\dfrac{5}{9}}\\\\\sin\alpha=\pm\dfrac{\sqrt5}{3}\\\\\huge\boxed{\sin\alpha=-\dfrac{\sqrt5}{3}}[/tex]
[tex]\text{tg}\alpha=\dfrac{-\frac{\sqrt5}{3}}{\frac{2}{3}}=-\dfrac{\sqrt5}{3}\cdot\dfrac{3}{2}\\\\\huge\boxed{\text{tg}\alpha=-\dfrac{\sqrt5}{2}}[/tex]
[tex]\text{ctg}\alpha=-\dfrac{2}{\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}\\\\\huge\boxed{\text{ctg}\alpha=-\dfrac{2\sqrt5}{5}}[/tex]
