Если tgA=1/2 то найдите

Решите задачу:

$\sin(2\alpha +\frac{\pi}{4} )=\sin2\alpha \cos\frac{\pi}{4}+\cos2\alpha \sin\frac{\pi}{4}=\sin2\alpha \cdot\frac{\sqrt{2}}{2}+\cos2\alpha \cdot\frac{\sqrt{2}}{2}= \\ \\ =\frac{\sqrt{2}}{2}\cdot(2\sin\alpha \cos\alpha +\cos^2\alpha -\sin^2\alpha )=\\\\=\frac{\sqrt{2}}{2}\cdot\cos^2\alpha\cdot(\frac{2\sin\alpha \cos\alpha}{\cos^2\alpha}+1 -\frac{\sin^2\alpha }{\cos^2\alpha})=$

$=\frac{\sqrt{2}}{2}\cdot\frac{1}{\frac{1}{\cos^2\alpha}}\cdot(\frac{2\sin\alpha }{\cos\alpha}+1 -tg^2\alpha)= \frac{\sqrt{2}}{2}\cdot\frac{1}{1+tg^2\alpha}\cdot(2tg\alpha+1 -tg^2\alpha)= \\ \\ = \frac{\sqrt{2}}{2}\cdot\frac{1}{1+(\frac{1}{2})^2}\cdot(2\cdot\frac{1}{2}+1- (\frac{1}{2})^2)= \frac{\sqrt{2}}{2}\cdot\frac{1}{1+\frac{1}{4}}\cdot(1+1- \frac{1}{4})= \\ \\ = \frac{\sqrt{2}}{2}\cdot\frac{1}{\frac{5}{4}}\cdot(2- \frac{1}{4})=\frac{\sqrt{2}}{2}\cdot\frac{4}{5}\cdot\frac{7}{4}= \frac{7\sqrt{2}}{10}$