Biết \( \sin x + \cos x = \dfrac{{ \sqrt 2 }}{2} \). Trong các kết quả sau, kết quả nào sai?
Giải chi tiết:
\(\begin{array}{l}\sin x + \cos x = \dfrac{{\sqrt 2 }}{2} \Rightarrow {\left( {\sin x + \cos x} \right)^2} = \dfrac{1}{2}\\ \Leftrightarrow {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = \dfrac{1}{2}\\ \Leftrightarrow 1 + 2\sin x\cos x = \dfrac{1}{2} \Leftrightarrow \sin x\cos x = - \dfrac{1}{4}\end{array}\)
Ta có: \({\left( {\sin x - \cos x} \right)^2} = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x\)
\( = 1 - 2.\left( { - \dfrac{1}{4}} \right) = \dfrac{3}{2} \Rightarrow \sin x - \cos x = \pm \dfrac{{\sqrt 6 }}{2}\).
\( \Rightarrow {\sin ^4}x + {\cos ^4}x = \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 2{\sin ^2}x{\cos ^2}x = 1 - 2.{\left( { - \dfrac{1}{4}} \right)^2} = \dfrac{7}{8}\).
\( \Rightarrow {\tan ^2}x + {\cot ^2}x = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = \dfrac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\sin }^2}x{{\cos }^2}x}} = \dfrac{{\dfrac{7}{8}}}{{{{\left( { - \dfrac{1}{4}} \right)}^2}}} = 14\).
Vậy khẳng định D sai.
Chọn D