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Câu hỏi: Cho \(\cos \alpha = \dfrac{1}{3}\), tính \(sin(\alpha + \dfrac{\pi }{6}) - \cos (\alpha - \dfrac{{2\pi }}{3})\)
Lời giải chi tiết
Ta có
\(sin(\alpha + \dfrac{\pi }{6}) - \cos (\alpha - \dfrac{{2\pi }}{3})\)
=\(sin\alpha c{\rm{os}}\dfrac{\pi }{6} + \cos \alpha \sin \dfrac{\pi }{6} - \cos \alpha \cos \dfrac{{2\pi }}{3} - \sin \alpha \sin \dfrac{{2\pi }}{3}\)
\(= \dfrac{{\sqrt 3 }}{2}sin\alpha + \dfrac{1}{2}\cos \alpha + \dfrac{1}{2}\cos \alpha - \dfrac{{\sqrt 3 }}{2}\sin \alpha \)
\(= \cos \alpha = \dfrac{1}{3}\)
Ta có
\(sin(\alpha + \dfrac{\pi }{6}) - \cos (\alpha - \dfrac{{2\pi }}{3})\)
=\(sin\alpha c{\rm{os}}\dfrac{\pi }{6} + \cos \alpha \sin \dfrac{\pi }{6} - \cos \alpha \cos \dfrac{{2\pi }}{3} - \sin \alpha \sin \dfrac{{2\pi }}{3}\)
\(= \dfrac{{\sqrt 3 }}{2}sin\alpha + \dfrac{1}{2}\cos \alpha + \dfrac{1}{2}\cos \alpha - \dfrac{{\sqrt 3 }}{2}\sin \alpha \)
\(= \cos \alpha = \dfrac{1}{3}\)