Problem 41

We're given that $\tan (\frac{π}{4}+α)=2$. To find $\sin 2α$, we'll use the trigonometric identity for the tangent of a sum of angles, which is:
$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$

Let $A = \frac{π}{4}$ and $B = α$. Then, $\tan \frac{π}{4} = 1$, and we can substitute these values into the identity:
$$2 = \tan (\frac{π}{4}+α) = \frac{1 + \tan α}{1 - \tan α}$$

Solving for $\tan α$, we get:
$$\tan α = \frac{2 - 1}{1 + 2} = \frac{1}{3}$$

Now, to find $\sin 2α$, we'll use the double-angle identity for sine:
$$\sin 2α = \frac{2\tan α}{1 + \tan^2 α}$$

Substituting $\tan α = \frac{1}{3}$ into the identity:
$$\sin 2α = \frac{2 \cdot \frac{1}{3}}{1 + (\frac{1}{3})^2} = \frac{\frac{2}{3}}{1 + \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{10}{9}} = \boxed{\frac{3}{5}}$$