Firing up a slope: the best angle

Origin of the formula (for completeness). With $x=vt\cos\theta$ and $y=vt\sin\theta-\frac12gt^2$, the projectile lands on the slope $y=x\tan\alpha$ when $v\sin\theta-\frac12gt=v\cos\theta\tan\alpha$, i.e. at $t=\frac{2v\sin(\theta-\alpha)}{g\cos\alpha}$. The distance along the slope is $\frac{x}{\cos\alpha}=\frac{vt\cos\theta}{\cos\alpha}$, which gives the formula above.

Simplification.
$$R(\theta)=\frac{v^2}{g\cos^2\alpha}\bigl(\sin(2\theta-\alpha)-\sin\alpha\bigr).$$
Only $\sin(2\theta-\alpha)$ depends on $\theta$.

Maximum. $R'(\theta)=\dfrac{2v^2}{g\cos^2\alpha}\cos(2\theta-\alpha)$ vanishes for $2\theta-\alpha=\frac\pi2$, i.e.
$$\theta=\frac\pi4+\frac\alpha2,$$
which lies in $\left(\alpha,\frac\pi2\right)$. On the interval, $2\theta-\alpha$ goes from $\alpha$ to $\pi-\alpha$ and $\sin$ is maximal at $\frac\pi2$. The optimal direction bisects the angle between the slope (direction $\alpha$) and the vertical (direction $\frac\pi2$).

Maximal range. With $\sin(2\theta-\alpha)=1$ and $\cos^2\alpha=(1-\sin\alpha)(1+\sin\alpha)$,
$$R_{\max}=\frac{v^2(1-\sin\alpha)}{g\cos^2\alpha}=\frac{v^2}{g(1+\sin\alpha)}.$$

Checks. $\alpha=0$ gives the classical $\theta=\frac\pi4$ and $R_{\max}=\frac{v^2}g$. Firing downhill (replace $\alpha$ by $-\alpha$) gives $\frac{v^2}{g(1-\sin\alpha)}$, larger, as expected.