Упрощение тригонометрических выражений. Задача 23

\(\dfrac{{\sin \left( {\dfrac{{5\pi }}{2} + \dfrac{\alpha }{2}} \right)\left( {1 + {\rm{t}}{{\rm{g}}^2}\left( {\dfrac{{3\alpha }}{4}-\dfrac{\pi }{2}} \right)} \right){{\cos }^2}\dfrac{\alpha }{4}}}{{{\rm{t}}{{\rm{g}}^2}\left( {\dfrac{{3\pi }}{2}-\dfrac{\alpha }{4}} \right)-{\rm{t}}{{\rm{g}}^2}\left( {\dfrac{{3\alpha }}{4}-\dfrac{{7\pi }}{2}} \right)}} = \dfrac{{\cos \dfrac{\alpha }{2}\left( {1 + {\rm{ct}}{{\rm{g}}^2}\dfrac{{3\alpha }}{4}} \right){{\cos }^2}\dfrac{\alpha }{4}}}{{{\rm{ct}}{{\rm{g}}^2}\dfrac{\alpha }{4}-{\rm{ct}}{{\rm{g}}^2}\dfrac{{3\alpha }}{4}}} = \)

\( = \dfrac{{\cos \dfrac{\alpha }{2}{{\cos }^2}\dfrac{\alpha }{4}}}{{{{\sin }^2}\dfrac{{3\alpha }}{4} \cdot \left( {{\rm{ctg}}\dfrac{\alpha }{4}-{\rm{ctg}}\dfrac{{3\alpha }}{4}} \right)\left( {{\rm{ctg}}\dfrac{\alpha }{4} + {\rm{ctg}}\dfrac{{3\alpha }}{4}} \right)}} = \dfrac{{\cos \dfrac{\alpha }{2}{{\cos }^2}\dfrac{\alpha }{4}}}{{{{\sin }^2}\dfrac{{3\alpha }}{4} \cdot \dfrac{{\sin \dfrac{\alpha }{2}}}{{\sin \dfrac{\alpha }{4}\sin \dfrac{{3\alpha }}{4}}} \cdot \dfrac{{\sin \alpha }}{{\sin \dfrac{\alpha }{4}\sin \dfrac{{3\alpha }}{4}}}}} = \)

\( = \dfrac{{\cos \dfrac{\alpha }{2} \cdot {{\cos }^2}\dfrac{\alpha }{4} \cdot {{\sin }^2}\dfrac{\alpha }{4}}}{{\sin \dfrac{\alpha }{2}\sin \alpha }} = \dfrac{{\cos \dfrac{\alpha }{2} \cdot \dfrac{1}{4} \cdot {{\left( {2\cos \dfrac{\alpha }{4}\sin \dfrac{\alpha }{4}} \right)}^2}}}{{\sin \dfrac{\alpha }{2} \cdot 2\sin \dfrac{\alpha }{2}\cos \dfrac{\alpha }{2}}} = \dfrac{{{{\sin }^2}\dfrac{\alpha }{2}}}{{8{{\sin }^2}\dfrac{\alpha }{2}}} = \dfrac{1}{8}.\)