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Решите задачу:

$\frac{sin(5\pi -\alpha )+cos(\frac{5\pi}{2}-5\alpha )+sin(3\alpha -\pi )}{2cos2\alpha-1} =\frac{sin(\pi -\alpha )+cos(\frac{\pi}{2}-5\alpha )-sin(\pi-3\alpha )}{2cos2\alpha-1} =$ $=\frac{sin\alpha+sin5\alpha -sin3\alpha}{2cos2\alpha-1} = \frac{2sin3\alpha*cos2\alpha -sin3\alpha}{2cos2\alpha-1} =\frac{sin3\alpha(2cos2\alpha -1)}{2cos2\alpha-1} = sin3\alpha$