I need to minimise the following function of $\alpha$ $$ f(\alpha) = (2\alpha - 1)A - 2\sqrt{\alpha(1-\alpha)}B $$ where $A$ and $B$ are independent of $\alpha$.
My Attempt
The derivative is $$ \frac{d}{d\alpha} f(\alpha) = 2A - \frac{1-2\alpha}{\sqrt{\alpha(1-\alpha)}}B $$ Setting this to zero we get $$ \frac{d}{d\alpha} f(\alpha) = 0 \implies 2\sqrt{\alpha(1-\alpha)}A = (1-2\alpha) B $$ Squaring both sides we get $$ (2\alpha - 2\alpha^2) A = (1 + 4\alpha^2 - 4\alpha) B $$ Rearranging we get $$ \alpha^2(2A + 4B) + \alpha(-2A-4B) + B = 0 $$ Using the quadratic formula $$ \begin{align} \alpha_{1, 2} &= \frac{2A + 4B \pm \sqrt{4A^2 + 16B^2 + 16 AB - 4B(2A + 4B)}}{2B} \\ &= \frac{2A + 4B \pm \sqrt{4A^2 + 8AB}}{2B} \end{align} $$ If I plug this into $f(\alpha)$ I don't get a nice expression and I am struggling with the second derivative test.
Second derivative
$$ \begin{align} \frac{d}{d\alpha^2} f(\alpha) &= -B\left(- \frac{2}{\sqrt{\alpha(1-\alpha)}} -\frac{(1-2\alpha)^2}{2\sqrt{\alpha^3(1-\alpha)^3}}\right) \end{align} $$
Following comment hint
Setting $\alpha = \cos^2\theta$ then we have $$ \begin{align} f(\theta) &= (2\cos^2\theta - 1) A - 2\sqrt{\cos^2\theta(1-\cos^2\theta)} \\ &= \cos(2\theta) A - 2\cos\theta \sin \theta B \\ &= \cos(2\theta) A - \sin(2\theta) B \end{align} $$ The derivative with respect to $\theta$ would then be $$ \frac{d}{d\theta} f(\theta) = -2\sin(2\theta) A -2\cos(2\theta) B $$ Setting this to zero gives $$ -\frac{\sin(2\theta^*)}{\cos(2\theta^*)} = \frac{B}{A} $$ which means $$ \theta^* = \frac{1}{2}\text{arc} \tan\left(-\frac{B}{A}\right) $$
Hint: Put $\alpha=\cos^{2}(\theta)$. The function becomes $A\cos (2\theta)-B\sin(2 \theta)$. Use C-S inequality to show that the minimum is $-\sqrt {A^{2}+B^{2}}$.