Bài 7: Cho \(\pi < \alpha < {{3\pi } \over 2}\). Xác định dấu của các giá trị lượng giác sau
a) \(\cos (\alpha – {\pi \over 2})\);
b) \(\sin ({\pi \over 2} + \alpha )\);
c) \(\tan ({{3\pi } \over 2} – \alpha )\);
d) \(\cot (\alpha + \pi )\)
a) Với \(\pi < \alpha < {{3\pi } \over 2}\) thì \({\pi \over 2} < \alpha – {\pi \over 2} < \pi \), do đó \(\cos (\alpha – {\pi \over 2}) < 0\).
b) \({{3\pi } \over 2} < {\pi \over 2} + \alpha < 2\pi \) nên \(\sin ({\pi \over 2} + \alpha ) < 0\)
c) \(0 < {{3\pi } \over 2} – \alpha < {\pi \over 2}\) nên \(\tan ({{3\pi } \over 2} – \alpha ) > 0\)
d) \(\pi < \alpha + \pi < {{5\pi } \over 2}\) nên \(\cot (\alpha + \pi ) > 0\)
Bài 8: Chứng minh rằng với mọi \(\alpha \), ta luôn có
a) \(\sin (\alpha + {\pi \over 2}) = \cos \alpha \);
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b) \({\rm{cos}}(\alpha + {\pi \over 2}) = – \sin \alpha \);
c) \(\tan (\alpha + {\pi \over 2}) = – \cot \alpha \);
d) \(\cot (\alpha + {\pi \over 2}) = – \tan \alpha \).
a) \(\sin (\alpha + {\pi \over 2}) = \sin ({\pi \over 2} – ( – \alpha )) = c{\rm{os( – }}\alpha {\rm{) = cos}}\alpha \)
b) \({\rm{cos}}(\alpha + {\pi \over 2}) = c{\rm{os(}}{\pi \over 2} – ( – \alpha ) = \sin ( – \alpha ) = – \sin \alpha \)
c) \(\tan (\alpha + {\pi \over 2}) = {{\sin (\alpha + {\pi \over 2})} \over {\cos (\alpha + {\pi \over 2})}} = {{\cos \alpha } \over { – \sin \alpha }} = – \cot \alpha \)
d) \(\cot (\alpha + {\pi \over 2}) = {{\cos (\alpha + {\pi \over 2})} \over {\sin (\alpha + {\pi \over 2})}} = {{ – \sin \alpha } \over {\cos \alpha }} = – \tan \alpha \)
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Bài 9: Tính các giá trị lượng giác của góc \(\alpha \), nếu
a) \({\rm{cos}}\alpha = – {1 \over 4},\pi < \alpha < {{3\pi } \over 2}\)
b) \({\rm{sin}}\alpha = {2 \over 3},{\pi \over 2} < \alpha < \pi \)
c) \({\rm{tan}}\alpha = {7 \over 3},0 < \alpha < {\pi \over 2}\)
d) \({\rm{cot}}\alpha = – {{14} \over 9},{{3\pi } \over 2} < \alpha < 2\pi \)
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 \(\sin \alpha = – {1 \over {\sqrt {1 + {{196} \over {81}}} }} = – {9 \over {\sqrt {277} }}\)
\({\rm{cos}}\alpha {\rm{ = sin}}\alpha {\rm{cot}}\alpha {\rm{ = }}{{14} \over {\sqrt {277} }},\tan \alpha = {1 \over {\cot \alpha }} = – {9 \over {14}}\)
