Đáp ána) Ta có:\(\sin ({\pi \over 3} + \alpha ) - \sin ({\pi \over 3} - \alpha ) = 2\cos {\pi \over 3}\sin \alpha = \sin \alpha \)b) Áp dụng: \({\cos ^2}a = {{1 + \cos 2a} \over 2}\) , ta có: \(\eqalign{
& {\cos ^2}({\pi \over 4} + \alpha ) - {\cos ^2}({\pi \over 4} - \alpha ) \cr&= {{1 + \cos ({\pi \over 2} + 2\alpha )} \over 2} - {{1 + \cos ({\pi \over 2} - 2\alpha )} \over 2} \cr
& = {1 \over 2}( - \sin 2\alpha - \sin 2\alpha ) = - \sin 2\alpha \cr} \)
