Bài tập 6.30 trang 189 SBT Toán 10

\(\begin{array}{l}
\sin \left( {\alpha  + \frac{\pi }{6}} \right) - \cos \left( {\alpha  - \frac{{2\pi }}{3}} \right)\\
 = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} - \cos \alpha \cos \frac{{2\pi }}{3} - \sin \alpha \sin \frac{{2\pi }}{3}\\
 = \frac{{\sqrt 3 }}{2}\sin \alpha  + \frac{1}{2}\cos \alpha  + \frac{1}{2}\cos \alpha  - \frac{{\sqrt 3 }}{2}\sin \alpha \\
 = \cos \alpha  = \frac{1}{3}
\end{array}\)

-- Mod Toán 10 HỌC247