$\alpha -\beta =\frac{\pi }{4}\Rightarrow \tan (\alpha -\beta )=\tan \frac{\pi }{4}$ $\Rightarrow \frac{\tan \alpha -\tan \beta }{1+\tan \alpha .\tan \beta }=1$ $\Rightarrow \frac{\tan \alpha -\frac{1}{2}}{1+\frac{1}{2}\tan \alpha }=1\Rightarrow \tan \alpha -\frac{1}{2}=1+\frac{1}{2}\tan \alpha $ $\Rightarrow \frac{1}{2}\tan \alpha =\frac{3}{2}\Rightarrow \tan \alpha =3$ با توجه به این که $\sin 2\alpha =2\sin \alpha \cos \alpha $، داریم: $\sin 2\alpha =2(\frac{3}{\sqrt{10}})(\frac{1}{\sqrt{10}})=\frac{6}{10}$