Cho biết π 2 < x < π và s i n ( x ) = 1 3 . Tính c o s ( x ) A. c o s ( x ) = 2 3

7/20

Cho biết \[\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < x < \pi }}\]và \[{\rm{sin}}\left( {\rm{x}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}\]. Tính\[{\rm{cos}}\left( {\rm{x}} \right)\]A. \[{\rm{cos}}\left( {\rm{x}} \right){\rm{ = }}\frac{{\rm{2}}}{{\rm{3}}}\]

\[{\rm{cos}}\left( {\rm{x}} \right){\rm{ = }} - \frac{2}{3}\]

\[{\rm{cos}}\left( {\rm{x}} \right){\rm{ = }}\frac{{{\rm{2}}\sqrt {\rm{2}} }}{{\rm{3}}}\]

\[{\rm{cos}}\left( {\rm{x}} \right){\rm{ = }} - \frac{{2\sqrt 2 }}{3}\]

Giải thích

Ta có:\[{\rm{si}}{{\rm{n}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ = 1}} \Rightarrow {\left( {\frac{{\rm{1}}}{{\rm{3}}}} \right)^{\rm{2}}}{\rm{ + co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ = 1}} \Rightarrow {\rm{co}}{{\rm{s}}^{\rm{2}}}\left( {\rm{x}} \right){\rm{ = 1}} - \frac{{\rm{1}}}{{\rm{9}}}{\rm{ = }}\frac{{\rm{8}}}{{\rm{9}}}\]

Mà\[\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ < x < \pi }} \Rightarrow {\rm{cos}}\left( {\rm{x}} \right){\rm{ < 0}}\]

\[ \Rightarrow {\rm{cos}}\left( {\rm{x}} \right){\rm{ = }} - \sqrt {\frac{{\rm{8}}}{{\rm{9}}}} {\rm{ = }} - \frac{{2\sqrt 2 }}{3}\]

Đáp án cần chọn là: D