How can one get the minimum of a trigonometric expression if the first-order derivative is not in a nice form?
The expression is this:
$$ \left[\cos ^{2}\left(\dfrac{\theta }{2}\right) + \,\sqrt{\,{1 - \gamma}\,}\,\sin^{2}\left(\dfrac{\theta }{2}\right)\right] ^{2} + \dfrac{\gamma }{4}\,\sin^{2}\left(\theta\right)\quad \mbox{where}\quad 0 \le \gamma \le 1. $$ I have tried to solve this but I reach a term which is not easily solvable.
Note $$\cos^2 \frac{\theta}{2} = \frac{1 + \cos \theta}{2}, \quad \sin^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{2},$$ and $$\sin^2 \theta = 1 - \cos^2 \theta.$$ So if we let $x = \cos \theta$, we get the function $$\begin{align}f(x) &= \left( \frac{1 + x}{2} + \sqrt{1-\gamma} \frac{1 - x}{2} \right)^2 + \frac{\gamma}{4} (1 - x^2) \\ &= ax^2 + bx + c, \end{align}$$ where $$\begin{align} a &= \frac{1 - \gamma - \sqrt{1-\gamma}}{2}, \\ b &= \frac{\gamma}{2}, \\ c &= \frac{1 + \sqrt{1-\gamma}}{2}. \end{align}$$ Now compute the critical point(s) of $f$ with respect to $x$ and consider where these have absolute value less than or equal to $1$, since we require $|x| = |\cos \theta| \le 1$.