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Решите задачу:

$\sin^2\bigg( \dfrac{\pi}{4} + \alpha \bigg)-\sin^2\bigg(\dfrac{\pi}{4} - \alpha \bigg)= \dfrac{1-\cos\bigg(\dfrac{\pi}{2} + \alpha \bigg)}{2} - \dfrac{1-\cos\bigg(\dfrac{\pi}{2}- \alpha \bigg) }{2} =$

$= \dfrac{1-\cos\bigg(\dfrac{\pi}{2}+ \alpha \bigg) -1+\cos\bigg(\dfrac{\pi}{2}- \alpha \bigg)}{2} = \dfrac{\sin \alpha +\sin \alpha }{2} =\sin \alpha$