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Решите задачу:

$\frac{1}{Sin\alpha+Sin3\alpha}+\frac{1}{Sin3\alpha+Sin5\alpha}=\frac{1}{2Sin\frac{\alpha+3\alpha}{2}Cos\frac{\alpha-3\alpha}{2}}+\frac{1}{2Sin\frac{3\alpha+5\alpha}{2} Cos\frac{3\alpha-5\alpha}{2} }=\frac{1}{2Sin2\alpha Cos\alpha}+\frac{1}{2Sin4\alpha Cos\alpha}=\frac{Sin4\alpha+Sin2\alpha}{2Sin2\alpha Sin4\alpha Cos\alpha}=\frac{2Sin3\alpha Cos\alpha}{2Sin2\alpha Sin4\alpha Cos\alpha}=\frac{Sin3\alpha }{Sin2\alpha Sin4\alpha}$

$\alpha=\frac{\pi }{12}\\\\\frac{Sin(3*\frac{\pi }{12}) }{Sin(2*\frac{\pi }{12})*Sin(4*\frac{\pi }{12})}=\frac{Sin\frac{\pi }{4} }{Sin\frac{\pi }{6}*Sin\frac{\pi }{3}}=\frac{\frac{\sqrt{2} }{2} }{\frac{1}{2}*\frac{\sqrt{3} }{2}}=\frac{\sqrt{2}*4 }{2*\sqrt{3} }=\frac{2\sqrt{2} }{\sqrt{3} }=\frac{2\sqrt{6} }{3}$