Projectile onto an inclined plane: The maximum range

Question:
A projectile is launched from the origin with initial velocity $v_0$ and the angle $\theta$ from the $x$-axis. The object is reached on the inclined plane below $x$-axis with angle of $\alpha$. Assume that the mass, initial velocity and angle $\alpha$ are fixed. Then, what is the angle $\theta$ to obtain maximum range on the inclined plane?

Answer:
The position of the projectile is given as
\begin{eqnarray}
x &=& v_0\cos\theta \ t  \\
y &=& v_0\sin\theta \ t - \frac{1}{2}gt^2
\end{eqnarray}
For the inclined plane, we can have the relationship between $x$ and $y$ as follows:
\begin{equation}
y = -x\tan\alpha
\end{equation}
Use (3) to eliminate $y$.
\begin{equation}
-x\tan\alpha = v_0 \sin\theta \ t - \frac{1}{2}gt^2
\end{equation}
From (1), we have $t = \frac{x}{v_0 \cos\theta}$ to eliminate $t$.
\begin{eqnarray}
-x\tan\alpha = v_0 \sin\theta\frac{x}{v_0 \cos\theta} - \frac{1}{2}g\left(\frac{x}{v_0 \cos\theta}\right)^2   \\
-x\tan\alpha = x\tan\theta - \frac{1}{2}\frac{gx^2}{v_0^2 \cos^2\theta} \\
x(\tan\alpha + \tan\theta) - \frac{1}{2}\frac{gx^2}{v_0^2 \cos^2\theta} = 0  \\
x \left\{ (\tan\alpha + \tan\theta) - \frac{1}{2}\frac{gx}{v_0^2 \cos^2\theta} \right\} = 0
\end{eqnarray}
$x=0$ is also the solution, but we are interested in the other one. Now, we can solve for $x$ as follows:
\begin{eqnarray}
x &=& \frac{2v_0^2\cos^2\theta (\tan\alpha + \tan\theta)}{g}\\
  &=& \frac{2v_0^2\cos^2\theta \left(\frac{\sin\alpha}{\cos\alpha} + \frac{\sin\theta}{\cos\theta} \right)}{g}\frac{\cos\alpha}{\cos\alpha}\\
 &=& \frac{2v_0^2\cos\theta (\sin\alpha\cos\theta + \sin\theta\cos\alpha)}{g\cos\alpha} \\
 &=& \frac{2v_0^2\cos\theta \sin(\alpha+\theta)}{g\cos\alpha} \\
 &=& \frac{2v_0^2 \left\{ \frac{1}{2}\sin(2\theta+\alpha)+\sin\alpha\right\}}{g\cos\alpha} \\
 &=& \frac{v_0^2 \left\{\sin(2\theta+\alpha)+\sin\alpha\right\}}{g\cos\alpha}
\end{eqnarray}
$v_0$ and $\alpha$ are constant, so the value of $\sin(2\theta+\alpha)$ affects the value of the whole function. A sine or cosine function oscillates between -1 and 1 assuming that the amplitude is 1. Thus, when we have
\begin{equation}
\sin(2\theta+\alpha) = 1
\end{equation}
the range becomes maximum. The angle must be
\begin{eqnarray}
2\theta+\alpha &=& \sin^{-1}1 = \frac{\pi}{2} \\
2\theta &=& \frac{\pi}{2}-\alpha  \\
\theta &=& \frac{\pi}{4} - \frac{\alpha}{2}
\end{eqnarray}

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