\sin{ \alpha} =\frac{4}{5}
Z jedynki trygonometrycznej:
sin^2 {\alpha}+\cos^2{\alpha}=1
Zatem
(\frac{4}{5})^2+\cos^2{\alpha}=1
\cos^2{\alpha}=1-\frac{16}{25}
\cos^2{\alpha}=\frac{9}{25}
Stąd:
1^0 \cos{\alpha}=\frac {3}{5} lub \cos{\alpha}=-\frac {3}{5}
\tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}}
\cot(\alpha)=\frac{1}{\tan{\alpha}}
Dla 1^0
\tan{\alpha}=\frac{\frac{4}{5}}{\frac {3}{5}}=\frac{4}{3}
\cot{\alpha}=\frac{1}{\frac{4}{3}}=\frac{3}{4}
Dla 2^0
\tan{\alpha}=\frac{\frac{4}{5}}{-\frac {3}{5}}=-\frac{4}{3}
\cot{\alpha}=\frac{1}{-\frac{4}{3}}=-\frac{3}{4}