Cho \(\cos \alpha = \frac{1}{3}\), tính \(\sin \left( {\alpha + \frac{\pi }{6}} \right) - \cos \left( {\alpha - \frac{{2\pi }}{3}} \right)\)
\(\begin{array}{l}
\sin \left( {\alpha + \frac{\pi }{6}} \right) - \cos \left( {\alpha - \frac{{2\pi }}{3}} \right)\\
= \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} - \cos \alpha \cos \frac{{2\pi }}{3} - \sin \alpha \sin \frac{{2\pi }}{3}\\
= \frac{{\sqrt 3 }}{2}\sin \alpha + \frac{1}{2}\cos \alpha + \frac{1}{2}\cos \alpha - \frac{{\sqrt 3 }}{2}\sin \alpha \\
= \cos \alpha = \frac{1}{3}
\end{array}\)
-- Mod Toán 10