Cho c o t ( α ) = 2 3 . Tính s i n ( 2 α + 7 π 4 )

Từ\[{\rm{cot}}\left( {\rm{\alpha }} \right){\rm{ = }}\frac{{\rm{2}}}{{\rm{3}}} \Rightarrow {\rm{tan}}\left( {\rm{\alpha }} \right){\rm{ = }}\frac{{\rm{3}}}{{\rm{2}}}\]

\[{\rm{sin}}\left( {{\rm{2\alpha }}} \right){\rm{ = }}\frac{{{\rm{2tan}}\left( {\rm{\alpha }} \right)}}{{{\rm{ta}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right){\rm{ + 1}}}}{\rm{ = }}\frac{{{\rm{2}}{\rm{.}}\frac{{\rm{3}}}{{\rm{2}}}}}{{{{\left( {\frac{{\rm{3}}}{{\rm{2}}}} \right)}^{\rm{2}}}{\rm{ + 1}}}}{\rm{ = }}\frac{{\rm{3}}}{{\frac{{{\rm{13}}}}{{\rm{4}}}}}{\rm{ = }}\frac{{{\rm{12}}}}{{{\rm{13}}}}\]

\[{\rm{cos}}\left( {{\rm{2\alpha }}} \right){\rm{ = }}\frac{{{\rm{1}} - {\rm{ta}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right)}}{{{\rm{1 + ta}}{{\rm{n}}^{\rm{2}}}\left( {\rm{\alpha }} \right)}}{\rm{ = }}\frac{{{\rm{1}} - {{\left( {\frac{{\rm{3}}}{{\rm{2}}}} \right)}^{\rm{2}}}}}{{{\rm{1 + }}{{\left( {\frac{{\rm{3}}}{{\rm{2}}}} \right)}^{\rm{2}}}}}{\rm{ = }}\frac{{ - {\rm{5}}}}{{{\rm{13}}}}\]

Ta có\[{\rm{sin}}\left( {{\rm{2\alpha + }}\frac{{{\rm{7\pi }}}}{{\rm{4}}}} \right){\rm{ = sin}}\left( {{\rm{2\alpha }} - \frac{{\rm{\pi }}}{{\rm{4}}}{\rm{ + 2\pi }}} \right){\rm{ = sin}}\left( {{\rm{2\alpha }} - \frac{{\rm{\pi }}}{{\rm{4}}}} \right)\]

\[{\rm{ = sin}}\left( {{\rm{2\alpha }}} \right){\rm{cos}}\left( {\frac{{\rm{\pi }}}{{\rm{4}}}} \right) - {\rm{cos}}\left( {{\rm{2\alpha }}} \right){\rm{sin}}\left( {\frac{{\rm{\pi }}}{{\rm{4}}}} \right){\rm{ = }}\frac{{{\rm{12}}}}{{{\rm{13}}}}{\rm{.}}\frac{{\sqrt {\rm{2}} }}{{\rm{2}}} - \left( {\frac{{ - {\rm{5}}}}{{{\rm{13}}}}} \right){\rm{.}}\frac{{\sqrt {\rm{2}} }}{{\rm{2}}}{\rm{ = }}\frac{{{\rm{17}}\sqrt {\rm{2}} }}{{{\rm{26}}}}\]

Đáp án cần chọn là: A