Gợi ý làm bài
a) \(\pi < \alpha < {{3\pi } \over 2} = > \sin \alpha < 0\)
Vậy \(\sin \alpha = - \sqrt {1 - {{\cos }^2}\alpha } = - \sqrt {1 - {1 \over {16}}} = - {{\sqrt {15} } \over 4}\)
\(\tan \alpha = {{\sin \alpha } \over {{\rm{cos}}\alpha }} = \sqrt {15} ,\cot \alpha = {1 \over {\sqrt {15} }}\)
b) \({\pi \over 2} < \alpha < \pi = > c{\rm{os}}\alpha {\rm{ < 0}}\)
Vậy \(\cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - {4 \over 9}} = {{ - \sqrt 5 } \over 3}\)
\(\tan \alpha = {{\sin \alpha } \over {c{\rm{os}}\alpha }}{\rm{ = - }}{2 \over {\sqrt 5 }}{\rm{,cot}}\alpha {\rm{ = - }}{{\sqrt 5 } \over 2}\)
c) \(0 < \alpha < {\pi \over 2} = \cos \alpha > 0,co{s^2}\alpha = {1 \over {1 + {{\tan }^2}\alpha }}\)
Vậy \(\cos \alpha = {1 \over {\sqrt {1 + {{49} \over 9}} }} = {3 \over {\sqrt {58} }}\)
\(\sin \alpha = \cos \alpha \tan \alpha = {7 \over {\sqrt {58} }},\cot \alpha = {3 \over 7}\)
d) \({{3\pi } \over 2} < \alpha < 2\pi = > \sin \alpha < 0,{\sin ^2}\alpha = {1 \over {1 + {{\cot }^2}\alpha }}\)
Vậy \(\sin \alpha = - {1 \over {\sqrt {1 + {{196} \over {81}}} }} = - {9 \over {\sqrt {277} }}\)
\({\rm{cos}}\alpha {\rm{ = sin}}\alpha {\rm{cot}}\alpha {\rm{ = }}{{14} \over {\sqrt {277} }},\tan \alpha = {1 \over {\cot \alpha }} = - {9 \over {14}}\)
