Gợi ý làm bài
a) \(0 < \alpha < {\pi \over 2} = > \cos \alpha > 0\), do đó
\(\cos \alpha = \sqrt {1 - si{n^2}\alpha } = \sqrt {1 - 0,36} = \sqrt {0,64} = 0,8\)
=> \(\tan \alpha = {3 \over 4},\cot \alpha = {4 \over 3}\)
b) \({\pi \over 2} < \alpha < \pi = > \sin \alpha > 0\), do đó
\(\sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } = \sqrt {1 - 0,49} = \sqrt {0,51} \approx 0,71\)
Suy ra: \(\tan \alpha = - {{0,7} \over {0,71}} \approx - 0,98,\cot \alpha \approx - 1,01\)
c) \(\pi < \alpha < {{3\pi } \over 2} = > \cos \alpha < 0\), do đó
\(\eqalign{
& \cos \alpha = - {1 \over {\sqrt {1 + {{\tan }^2}\alpha } }} = - {1 \over {\sqrt 5 }} = - {{\sqrt 5 } \over 5}, \cr
& \sin \alpha = - {{2\sqrt 5 } \over 5},\cot \alpha = {1 \over 2} \cr} \)
d) \({{3\pi } \over 2} < \alpha < 2\pi = > \sin \alpha < 0\), do đó
\(\eqalign{
& \sin \alpha = - {1 \over {\sqrt {1 + {{\cot }^2}\alpha } }} = - {1 \over {\sqrt {10} }} = - {{\sqrt {10} } \over {10}}, \cr
& cos\alpha = {{3\sqrt {10} } \over {10}},tan\alpha = - {1 \over 3} \cr} \)
