Cho \(\cos x = - \dfrac{1}{3};\left( {\dfrac{\pi }{3} < x < \pi } \right).\) Tính \(\sin \left( {x -
Cho \(\cos x = - \dfrac{1}{3};\left( {\dfrac{\pi }{3} < x < \pi } \right).\) Tính \(\sin \left( {x - \dfrac{\pi }{3}} \right)\)
+ Ta có: \({\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x = \dfrac{8}{9}\)
+ Do \(\dfrac{\pi }{3} < x < \pi \Rightarrow \sin x > 0 \Rightarrow \sin x = \sqrt {\dfrac{8}{9}} = \dfrac{{2\sqrt 2 }}{3}\)
+ Khi đó: \(\sin \left( {x - \dfrac{\pi }{3}} \right) = \sin x\cos \dfrac{\pi }{3} - \cos x\sin \dfrac{\pi }{3} = \dfrac{{2\sqrt 2 }}{3}.\dfrac{1}{2} - \left( { - \dfrac{1}{3}} \right).\dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt 3 + 2\sqrt 2 }}{6}\)
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