Cho cosa=1/8. Tính giá trị sin²a – cos²a /2.sina.cosa – cota 30/09/2021 Bởi Lyla Cho cosa=1/8. Tính giá trị sin²a – cos²a /2.sina.cosa – cota
Đáp án: \({\left[ {\begin{array}{*{20}{l}}{A = \dfrac{{10}}{{\sqrt 7 }}}\\{A = – \dfrac{{10}}{{\sqrt 7 }}{\rm{\;}}}\end{array}} \right.}\) Giải thích các bước giải: \(\begin{array}{*{20}{l}}{A = \dfrac{{{{\sin }^2}a – {{\cos }^2}a}}{{2\sin a\cos a}} – \cot a}\\{ = \dfrac{{1 – {{\cos }^2}a – {{\cos }^2}a}}{{2\sin a\cos a}} – \dfrac{{\cos a}}{{\sin a}}}\\{ = \dfrac{{1 – 2{{\cos }^2}a – 2{{\cos }^2}a}}{{2\sin a\cos a}}}\\{ = \dfrac{{1 – 4{{\cos }^2}a}}{{2\sin a\cos a}}}\\{Do:\cos a = \dfrac{1}{8}}\\{{{\cos }^2}a + {{\sin }^2}a = 1}\\{ \to \dfrac{1}{{64}} + {{\sin }^2}a = 1}\\{ \to {{\sin }^2}a = \dfrac{{63}}{{64}}}\\{ \to \left[ {\begin{array}{*{20}{l}}{\sin a = \dfrac{{\sqrt {63} }}{8}}\\{\sin a = – \dfrac{{\sqrt {63} }}{8}}\end{array}} \right.}\\{ \to \left[ {\begin{array}{*{20}{l}}{\dfrac{{1 – 4{{\cos }^2}a}}{{2\sin a\cos a}} = \dfrac{{1 – 4.\dfrac{1}{{64}}}}{{2.\dfrac{{\sqrt {63} }}{8}.\dfrac{1}{8}}}}\\{\dfrac{{1 – 4{{\cos }^2}a}}{{2\sin a\cos a}} = \dfrac{{1 – 4.\dfrac{1}{{64}}}}{{2.\left( { – \dfrac{{\sqrt {63} }}{8}} \right).\dfrac{1}{8}}}{\rm{\;}}}\end{array}} \right.}\\{ \to \left[ {\begin{array}{*{20}{l}}{A = \dfrac{{10}}{{\sqrt 7 }}}\\{A = – \dfrac{{10}}{{\sqrt 7 }}{\rm{\;}}}\end{array}} \right.}\end{array}\) Bình luận
Đáp án:
\({\left[ {\begin{array}{*{20}{l}}
{A = \dfrac{{10}}{{\sqrt 7 }}}\\
{A = – \dfrac{{10}}{{\sqrt 7 }}{\rm{\;}}}
\end{array}} \right.}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{A = \dfrac{{{{\sin }^2}a – {{\cos }^2}a}}{{2\sin a\cos a}} – \cot a}\\
{ = \dfrac{{1 – {{\cos }^2}a – {{\cos }^2}a}}{{2\sin a\cos a}} – \dfrac{{\cos a}}{{\sin a}}}\\
{ = \dfrac{{1 – 2{{\cos }^2}a – 2{{\cos }^2}a}}{{2\sin a\cos a}}}\\
{ = \dfrac{{1 – 4{{\cos }^2}a}}{{2\sin a\cos a}}}\\
{Do:\cos a = \dfrac{1}{8}}\\
{{{\cos }^2}a + {{\sin }^2}a = 1}\\
{ \to \dfrac{1}{{64}} + {{\sin }^2}a = 1}\\
{ \to {{\sin }^2}a = \dfrac{{63}}{{64}}}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\sin a = \dfrac{{\sqrt {63} }}{8}}\\
{\sin a = – \dfrac{{\sqrt {63} }}{8}}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\dfrac{{1 – 4{{\cos }^2}a}}{{2\sin a\cos a}} = \dfrac{{1 – 4.\dfrac{1}{{64}}}}{{2.\dfrac{{\sqrt {63} }}{8}.\dfrac{1}{8}}}}\\
{\dfrac{{1 – 4{{\cos }^2}a}}{{2\sin a\cos a}} = \dfrac{{1 – 4.\dfrac{1}{{64}}}}{{2.\left( { – \dfrac{{\sqrt {63} }}{8}} \right).\dfrac{1}{8}}}{\rm{\;}}}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{A = \dfrac{{10}}{{\sqrt 7 }}}\\
{A = – \dfrac{{10}}{{\sqrt 7 }}{\rm{\;}}}
\end{array}} \right.}
\end{array}\)
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