$$\frac{\sin 4\alpha }{4{{\sin }^{2}}\left( \frac{ \pi }{4}+\alpha \right)\tan \left( \frac{ \pi }{4}-\alpha \right)}=\frac{\sin 4\alpha }{4\sin \left( \frac{ \pi }{4}+\alpha \right)\sin \left( \frac{ \pi }{4}-\alpha \right)}=\frac{\sin 4\alpha }{2\sin \left( \frac{ \pi }{2}+2\alpha \right)}=\sin 2\alpha $$ 
化简$$\frac{\sin 4\alpha }{4{{\sin }^{2}}\left( \frac{ \pi }{4}+\alpha \right)\tan \left( \frac{ \pi }{4}-\alpha \right)}$$=．
(a) $$\cos 2\alpha $$ 
(b) $$\sin 2\alpha $$ 
(c) $$\cos \alpha $$ 
(d) $$\sin \alpha $$ 