The Theta-Beta-Mach relation, shown below, relates the shock wave angle $\beta$ to the flow deflection angle (or body deflection angle) $\theta$ in terms of the ratio of specific heats $\gamma$ and the pre-shock Mach Number $M_1$.
$$\tan \theta =
2\cot\beta\frac{M_1^2\sin^2\beta-1}{M_1^2(\gamma+\cos2\beta)+2}$$
As an example, this relation plots as such:
Shock Wave Angle Chart
Finding the maximum deflection angle $\theta_{\mathrm{max}}$ for given pre-shock conditions is of critical importance, and can be attempted by taking the implicit derivative of the Theta-Beta-Mach Relation with respect to $\theta$.
$$\frac{d}{d \theta} \left[ \tan \theta =
2\cot\beta\frac{M_1^2\sin^2\beta-1}{M_1^2(\gamma+\cos2\beta)+2} \right]$$
I will not be attempting this by hand, but have used a solver (https://www.derivative-calculator.net/#) to obtain the following:
$$ \frac{d \beta}{d \theta} =\dfrac{\left(M_1^2\cos\left(2{\beta}\right)+M_1^2{\gamma}+2\right)^3\sec^2\left({\theta}\right)}{\left(\left(4M_1^5\cot\left({\beta}\right)\sin^2\left({\beta}\right)-4M_1^4\cot\left({\beta}\right)\right)\cos\left(2{\beta}\right)+\left(4M_1^5{\gamma}+8M_1^3\right)\cot\left({\beta}\right)\sin^2\left({\beta}\right)+\left(-4M_1^4{\gamma}-8M_1^2\right)\cot\left({\beta}\right)\right)\sin\left(2{\beta}\right)-2M_1\csc^2\left({\beta}\right)\sin^2\left({\beta}\right)\left(M_1^2\cos\left(2{\beta}\right)+M_1^2{\gamma}+2\right)^2+4M_1\cos\left({\beta}\right)\cot\left({\beta}\right)\sin\left({\beta}\right)\left(M_1^2\cos\left(2{\beta}\right)+M_1^2{\gamma}+2\right)^2+2\csc^2\left({\beta}\right)\left(M_1^2\cos\left(2{\beta}\right)+M_1^2{\gamma}+2\right)^2} $$
Since we are trying to find where $\frac{d \beta}{d \theta} = 0$, the denominator can be entirely discounted, and we will instead look for where
$$ \left(M_1^2\cos\left(2{\beta}\right)+M_1^2{\gamma}+2\right)^3\sec^2\left({\theta}\right) = 0$$
However, neither term will ever equal zero for known conditions $1 < \gamma$, $1 < M_1$. How can $\theta_{\mathrm{max}}$ then be found analytically?
EDIT:
@TedShifrin, yes, I was thinking about this backwards. I should instead be finding the zero for the derivative $\frac{d \theta}{d \beta}$ which is the much more complex:
$$\frac{d \theta}{d\beta}=\dfrac{\left(\left(4M^6\cot\left({\beta}\right)\sin^2\left({\beta}\right)-4M^4\cot\left({\beta}\right)\right)\cos\left(2{\beta}\right)+\left(4M^6{\gamma}+8M^4\right)\cot\left({\beta}\right)\sin^2\left({\beta}\right)+\left(-4M^4{\gamma}-8M^2\right)\cot\left({\beta}\right)\right)\sin\left(2{\beta}\right)-2M^2\csc^2\left({\beta}\right)\sin^2\left({\beta}\right)\left(M^2\cos\left(2{\beta}\right)+M^2{\gamma}+2\right)^2+4M^2\cos\left({\beta}\right)\cot\left({\beta}\right)\sin\left({\beta}\right)\left(M^2\cos\left(2{\beta}\right)+M^2{\gamma}+2\right)^2+2\csc^2\left({\beta}\right)\left(M^2\cos\left(2{\beta}\right)+M^2{\gamma}+2\right)^2}{\left(M^2\cos\left(2{\beta}\right)\left(M^2\cos\left(2{\beta}\right)+M^2{\gamma}+2\right)^2+M^2{\gamma}\cdot\left(M^2\cos\left(2{\beta}\right)+M^2{\gamma}+2\right)^2+2\left(M^2\cos\left(2{\beta}\right)+M^2{\gamma}+2\right)^2\right)\sec^2\left({\theta}\right)}$$
A numerical solution will be much easier to employ.
Best Answer
Setting $M_1>1$, $\gamma>1$ and given the function $\theta:[\arcsin(1/M_1),\,\pi/2] \to \mathbb{R}$ of law:
$$ \theta(\beta) := \arctan\left(2\cot\beta\,\frac{M_1^2\sin^2\beta-1}{M_1^2\left(\gamma+\cos(2\,\beta)\right)+2}\right) $$
it follows that:
$$ \frac{\text{d}\theta}{\text{d}\beta} = 0 \quad \quad \quad \Leftrightarrow \quad \quad \quad \frac{\text{d}}{\text{d}\beta}\left(2\cot\beta\,\frac{M_1^2\sin^2\beta-1}{M_1^2\left(\gamma+\cos(2\,\beta)\right)+2}\right) = 0 $$
i.e.
$$ \frac{2\left[M_1^2\left(M_1^2\left(1+\gamma\cos(2\,\beta)\right)-4\right)\sin^2\beta+M_1^2\left(1+\gamma\right)+2\right]}{\left(M_1^2\left(\gamma+\cos(2\,\beta)\right)+2\right)^2\sin^2\beta} = 0 $$
i.e.
$$ M_1^2\left(M_1^2\left(1+\gamma\left(1-2\sin^2\beta\right)\right)-4\right)\sin^2\beta+M_1^2\left(1+\gamma\right)+2 = 0 $$
i.e.
$$ \sin^4\beta - \frac{M_1^2(1+\gamma)-4}{2\,M_1^2\,\gamma}\,\sin^2\beta - \frac{M_1^2(1+\gamma)+2}{2\,M_1^4\,\gamma} = 0 $$
from which what is desired:
$$ \boxed{\beta = \arcsin\left(\sqrt{\frac{\frac{M_1^2(1+\gamma)-4}{2\,M_1^2\,\gamma}+\sqrt{\left(\frac{M_1^2(1+\gamma)-4}{2\,M_1^2\,\gamma}\right)^2+4\left(\frac{M_1^2(1+\gamma)+2}{2\,M_1^4\,\gamma}\right)}}{2}}\right)}\,. $$
ADDENDUM:
Set:
$$ a \equiv \frac{M_1^2(1+\gamma)-4}{2\,M_1^2\,\gamma}, \quad \quad \quad b \equiv \frac{M_1^2(1+\gamma)+2}{2\,M_1^4\,\gamma}, \quad \quad \quad c \equiv a+\sqrt{a^2+4\,b} $$
then:
$$ \beta^* = \arcsin\left(\sqrt{\frac{c}{2}}\,\right), \quad \quad \quad \boxed{\theta_{\max} = \theta(\beta^*) = \arctan\left(\sqrt{\frac{2}{c}-1}\;\frac{M_1^2\,c-2}{M_1^2\,(1+\gamma-c)+2}\right)}\,. $$