Cho \({\rm{tan}}\alpha = \dfrac{{21}}{{20}}\) và \({180^ \circ } < \alpha < {270^ \circ }\). Tính \({\rm{cot}}\alpha ,{\rm{cos}}\alpha ,{\rm{sin}}\alpha \).
Câu 704466: Cho \({\rm{tan}}\alpha = \dfrac{{21}}{{20}}\) và \({180^ \circ } < \alpha < {270^ \circ }\). Tính \({\rm{cot}}\alpha ,{\rm{cos}}\alpha ,{\rm{sin}}\alpha \).
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Giải chi tiết:
Ta có: \({\rm{tan}}\alpha = \dfrac{{21}}{{20}} \Rightarrow {\rm{cot}}\alpha = \dfrac{1}{{{\rm{tan}}\alpha }} = \dfrac{{20}}{{21}}\)
Lại có: \({\rm{ta}}{{\rm{n}}^2}\alpha + 1 = \dfrac{1}{{{\rm{co}}{{\rm{s}}^2}\alpha }}\)
\( \Leftrightarrow {\rm{co}}{{\rm{s}}^2}\alpha = \dfrac{1}{{{\rm{ta}}{{\rm{n}}^2}\alpha + 1}} \Leftrightarrow {\rm{co}}{{\rm{s}}^2}\alpha = \dfrac{1}{{{{\left( {\dfrac{{21}}{{20}}} \right)}^2} + 1}} = \dfrac{{400}}{{841}} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{{\rm{cos}}\alpha = \dfrac{{20}}{{29}}}\\{{\rm{cos}}\alpha = - \dfrac{{20}}{{29}}}\end{array}} \right.\)
Do \({180^ \circ } < \alpha < {270^ \circ } \Rightarrow - 1 < {\rm{cos}}\alpha < 0\).
Suy ra: \({\rm{cos}}\alpha = - \dfrac{{20}}{{29}}\)
Khi đó: \({\rm{sin}}\alpha = {\rm{tan}}\alpha \cdot {\rm{cos}}\alpha = \dfrac{{21}}{{20}} \cdot \left( { - \dfrac{{20}}{{29}}} \right) = - \dfrac{{21}}{{29}}\)
Vậy \({\rm{cot}}\alpha = \dfrac{{20}}{{21}},{\rm{cos}}\alpha = - \dfrac{{20}}{{29}},{\rm{sin}}\alpha = - \dfrac{{21}}{{29}}\)
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