Cho \(\sin x = \dfrac{3}{5}\) biết \(\dfrac{\pi }{2} < x < \pi \) .Tính \(\sin \left( {2x + \dfrac{\pi }{6}}
Cho \(\sin x = \dfrac{3}{5}\) biết \(\dfrac{\pi }{2} < x < \pi \) .Tính \(\sin \left( {2x + \dfrac{\pi }{6}} \right)\)
+ Ta có: \({\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x = \dfrac{{16}}{{25}}\)
+ Do \(\dfrac{\pi }{2} < x < \pi \Rightarrow \cos x < 0 \Rightarrow \cos x = - \sqrt {\dfrac{{16}}{{25}}} = - \dfrac{4}{5}\)
+ Khi đó: \(\sin \left( {2x + \dfrac{\pi }{6}} \right) = \sin 2x\cos \dfrac{\pi }{6} + \cos 2x\sin \dfrac{\pi }{6} = 2\sin x\cos x\cos \dfrac{\pi }{6} + \left( {2{{\cos }^2}x - 1} \right)\sin \dfrac{\pi }{6}\)
\( = 2.\dfrac{3}{5}.\left( { - \dfrac{4}{5}} \right).\dfrac{{\sqrt 3 }}{2} + \left[ {2.{{\left( { - \dfrac{4}{5}} \right)}^2} - 1} \right].\dfrac{1}{2} = \dfrac{{7 - 24\sqrt 3 }}{{50}}\)
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