Đáp án:
\(\begin{array}{l}
\sin x = \dfrac{{2\sqrt 2 }}{3}\\
\sin \left( {x - \dfrac{\pi }{3}} \right) = \dfrac{{2\sqrt 2 + \sqrt 3 }}{6}\\
\cos \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{4 - \sqrt 2 }}{6}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{\pi }{2} < x < \pi \Rightarrow \sin x > 0\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x + {\left( { - \dfrac{1}{3}} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}x + \dfrac{1}{9} = 1\\
\Leftrightarrow {\sin ^2}x = \dfrac{8}{9}\\
\sin x > 0 \Rightarrow \sin x = \dfrac{{2\sqrt 2 }}{3}\\
\sin \left( {x - \dfrac{\pi }{3}} \right) = \sin x.\cos \dfrac{\pi }{3} - \cos x.\sin \dfrac{\pi }{3}\\
= \dfrac{{2\sqrt 2 }}{3}.\dfrac{1}{2} - \dfrac{{ - 1}}{3}.\dfrac{{\sqrt 3 }}{2} = \dfrac{{2\sqrt 2 + \sqrt 3 }}{6}\\
\cos \left( {x - \dfrac{\pi }{4}} \right) = \cos x.\cos \dfrac{\pi }{4} + \sin x.\sin \dfrac{\pi }{4}\\
= \dfrac{{ - 1}}{3}.\dfrac{{\sqrt 2 }}{2} + \dfrac{{2\sqrt 2 }}{3}.\dfrac{{\sqrt 2 }}{2} = \dfrac{{4 - \sqrt 2 }}{6}
\end{array}\)
