a) \(\sin \alpha = {4 \over 5}\,\,;\,\,\,\cos \alpha < 0\)
b) \(\cos \alpha = - {8 \over {17}};\,\,\,{\pi \over 2} < \alpha < \pi \)
c) \(\tan \alpha = \sqrt 3 \,\,;\,\,\,\pi < \alpha < {{3\pi } \over 2}\)
a) Ta có:
\(\eqalign{
& \cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \sqrt {1 - {{16} \over {25}}} = - {3 \over 5} \cr
& \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = - {4 \over 3} \cr
& \cot \alpha = {1 \over {\tan \alpha }} = - {3 \over 4} \cr} \)
b) Ta có:
\(\eqalign{
& \,{\pi \over 2} < \alpha < \pi \Rightarrow \sin \alpha > 0 \cr
& \Rightarrow \sin \alpha = \sqrt {1 - {{\cos }^2}\alpha } = \sqrt {1 - {{({8 \over {17}})}^2}} = {{15} \over {17}} \cr
& \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = - {{15} \over 8} \cr
& \cot \alpha = {1 \over {\tan \alpha }} = - {8 \over {15}} \cr} \)
c) Ta có:
\(\eqalign{
& \pi < \alpha < {{3\pi } \over 2} \Rightarrow \cos \alpha < 0 \cr
& \Rightarrow \cos \alpha = {{ - 1} \over {\sqrt {1 + {{\tan }^2}\alpha } }} = {{ - 1} \over {\sqrt {1 + {{(\sqrt 3 )}^2}} }} = - {1 \over 2} \cr
& \sin \alpha = - {{\sqrt 3 } \over 2} \cr
& \cot \alpha = {{\sqrt 3 } \over 3} \cr} \)
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