Đơn giản các biểu thức sau:
a) \(\sin \left( {\frac{\pi }{3} + \alpha } \right) - \sin \left( {\frac{\pi }{3} - \alpha } \right)\)
b) \({\cos ^2}\left( {\frac{\pi }{4} + \alpha } \right) - {\cos ^2}\left( {\frac{\pi }{4} - \alpha } \right)\)
a) Ta có
\(\begin{array}{l}
\sin \left( {\frac{\pi }{3} + \alpha } \right) - \sin \left( {\frac{\pi }{3} - \alpha } \right)\\
= 2\cos \frac{\pi }{3}\sin \alpha = \sin \alpha
\end{array}\)
b)
\(\begin{array}{l}
{\cos ^2}\left( {\frac{\pi }{4} + \alpha } \right) - {\cos ^2}\left( {\frac{\pi }{4} - \alpha } \right)\\
= \frac{{1 + \cos \left( {\frac{\pi }{2} + 2\alpha } \right)}}{2} - \frac{{1 + \cos \left( {\frac{\pi }{2} - 2\alpha } \right)}}{2}\\
= \frac{1}{2}\left( { - \sin 2\alpha - \sin 2\alpha } \right) = - \sin 2\alpha
\end{array}\)
-- Mod Toán 10
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