已知 6tan(α+π4)cos(α−π4)=1\sqrt{6}\tan \left( {\alpha + \frac{\pi }{4}}\right) \cos \left( {\alpha - \frac{\pi }{4}}\right) = 16tan(α+4π)cos(α−4π)=1 ，则 sin2α=\sin {2\alpha } =sin2α= （ ）A. −23- \frac{2}{3}−32B. −13- \frac{1}{3}−31C. 13\frac{1}{3}31D. 23\frac{2}{3}32