Transforming from one spherical coordinate system to another

To make things simpler I assume $\alpha_0=0$ (the desired value of $\alpha_0$ can always be added/subtracted at the end); furthermore I take the equator at $\delta=0$, so the range of the latitude $\delta$ is the interval $[-\pi/2,\pi/2]$. Your change of coordinates amounts to changing the standard basis $(e_1,e_, e_3)$ of ${\mathbb R}^3$ to the new basis

$$\bar e_1=(\cos\delta_0,0,\sin\delta_0), \quad \bar e_2=(0,1,0),\quad \bar e_3=(-\sin\delta_0,0,\cos\delta_0)\ .$$

It follows that the new coordinates $\bar x_k$ are given in terms of the old ones $x_i$ by the formulas

$$\bar x_1=\cos\delta_0 x_1 +\sin\delta_0 x_3, \quad \bar x_2=x_2,\quad \bar x_3=-\sin\delta_0 x_1+\cos\delta_0 x_3\ .$$

Now we have to express this in terms of the "geographical" quantities $\alpha$, $\delta$, resp. $\bar\alpha$, $\bar\delta$. On the one hand we have

$$x_1=\cos\delta\cos\alpha,\quad x_2=\cos\delta\sin\alpha, \quad x_3=\sin\delta\ ,$$

and on the other hand

$$\bar\alpha=\arg\Bigl({\bar x_1\over\rho},{\bar x_2\over\rho}\Bigr), \quad \bar\delta=\arcsin(\bar x_3)\ ,$$

where $\rho:=\sqrt{\bar x_1^2+\bar x_2^2}$. Putting it all together some simplifications will result.