Radius of the circumscribed circle of an isosceles triangle An isosceles triangle $ABC$ is given $(AC=BC).$ The perimeter of $\triangle ABC$ is $2p$, and the base angle is $\alpha.$ Find the radius of the circumscribed circle $R$.

$$R=\frac{p}{2\sin\alpha(1+\cos\alpha)}$$


Let $CD=2R.$ The triangle $BCD$ is a right triangle and we have $\angle BAC=\angle ABC=\angle BDC=\alpha.$
I am not sure how to approach the problem. It's really hard for me to solve problems like this. Can you give me a hint and some thoughts on the problem?

Hint: Use following formula:
$$R=\frac{p}{4\cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right)\cos\left(\frac{\gamma}{2}\right)}$$
Where $\alpha$, $\beta$ and $\gamma$ are angles on vertices A, B and C respectively. $\alpha=\beta$ , therefore we have:
$$R=\frac{p}{4\cos^2\left(\frac{\alpha}{2}\right)\cos\left(\frac{\gamma}{2}\right)}$$
And also:
$$2\alpha+\gamma=\pi$$
$$\implies\frac{\gamma}{2}=\frac{\pi}{2}-\frac{\alpha}{4}$$
Finally:
$$R=\frac{p}{4\cos^2\left(\frac{\alpha}{2}\right)\sin\left(\frac{\alpha}{4}\right)}$$
