=\[sin\alpha c{\rm{os}}\dfrac{\pi }{6} + \cos \alpha \sin \dfrac{\pi }{6} - \cos \alpha \cos \dfrac{{2\pi }}{3} - \sin \alpha \sin \dfrac{{2\pi }}{3}\]
Đề bài
Cho \[\cos \alpha = \dfrac{1}{3}\], tính \[sin[\alpha + \dfrac{\pi }{6}] - \cos [\alpha - \dfrac{{2\pi }}{3}]\]
Lời giải chi tiết
Ta có
\[sin[\alpha + \dfrac{\pi }{6}] - \cos [\alpha - \dfrac{{2\pi }}{3}]\]
=\[sin\alpha c{\rm{os}}\dfrac{\pi }{6} + \cos \alpha \sin \dfrac{\pi }{6} - \cos \alpha \cos \dfrac{{2\pi }}{3} - \sin \alpha \sin \dfrac{{2\pi }}{3}\]
\[ = \dfrac{{\sqrt 3 }}{2}sin\alpha + \dfrac{1}{2}\cos \alpha + \dfrac{1}{2}\cos \alpha - \dfrac{{\sqrt 3 }}{2}\sin \alpha \]
\[ = \cos \alpha = \dfrac{1}{3}\]