Cho sin x + cos x = 1/3 . Tính cos 4x .

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Cho\(\sin x + \cos x = \frac{1}{3}\). Tính \(\cos 4x\).

\(\cos 4x = \frac{{25}}{9}\).

\(\cos 4x = \frac{{209}}{{81}}\).

\(\cos 4x = - \frac{{47}}{{81}}\).

\(\cos 4x = - \frac{8}{9}\).

Giải thích

Chọn C

Ta có:

\({\left( {\sin x + \cos x} \right)^2} = \frac{1}{9} \Leftrightarrow {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x = \frac{1}{9} \Leftrightarrow \sin 2x = \frac{1}{9} - 1 \Leftrightarrow \sin 2x = - \frac{8}{9}\)

\(\cos 4x = 1 - 2{\sin ^2}2x = 1 - 2.{\left( { - \frac{8}{9}} \right)^2} = - \frac{{47}}{{81}}\)