The maximum range is determined by the total flight time, which is assumed constant. In still air, the time is $T = R/v$. Let the plane's ground track be the x-axis. The wind vector $\vec{u}$ makes an angle $(\beta-\alpha)$ with this axis. The component of wind velocity along the track is $u_\parallel = u \cos(\beta-\alpha)$, and perpendicular to the track is $u_\perp = u \sin(\beta-\alpha)$. For the plane to stay on course, its velocity relative to the air, $\vec{v}$, must compensate for the perpendicular wind component. The component of $\vec{v}$ along the track is $v_\parallel = \sqrt{v^2 - u_\perp^2} = \sqrt{v^2 - u^2\sin^2(\beta-\alpha)}$.

The ground speed on the outbound trip is $v_{out} = v_\parallel + u_\parallel$. The ground speed on the return trip is $v_{back} = v_\parallel - u_\parallel$.

Let the new one-way distance be $L$. The total time is:

$$T = t_{out} + t_{back} = \frac{L}{v_{out}} + \frac{L}{v_{back}} = L \left( \frac{1}{v_\parallel + u_\parallel} + \frac{1}{v_\parallel - u_\parallel} \right)$$ $$T = L \frac{2v_\parallel}{v_\parallel^2 - u_\parallel^2}$$

Substitute $v_\parallel^2 = v^2 - u_\perp^2$ and $u_\parallel^2 + u_\perp^2 = u^2$:

$$v_\parallel^2 - u_\parallel^2 = (v^2 - u_\perp^2) - u_\parallel^2 = v^2 - (u_\perp^2 + u_\parallel^2) = v^2 - u^2$$

So, $T = L \frac{2\sqrt{v^2 - u^2\sin^2(\beta-\alpha)}}{v^2 - u^2}$. Equating this to the still-air time $T=R/v$:

$$\frac{R}{v} = L \frac{2\sqrt{v^2 - u^2\sin^2(\beta-\alpha)}}{v^2 - u^2}$$

The new maximum round-trip range is $D = 2L$.

$$D = 2L = \frac{R}{v} \frac{v^2 - u^2}{\sqrt{v^2 - u^2\sin^2(\beta-\alpha)}}$$