Trig +Nested Radicals = Fun

We know that $\cos 2\theta = 2\cos^2 \theta - 1$ . Then $\cos \theta = 2\cos \frac \theta 2-1$ and

$\begin{aligned} \cos \frac \theta 2 & = \sqrt{\frac 12 + \frac 12 \cos \theta} \\ \cos \frac \theta 4 & = \sqrt{\frac 12 + \frac 12 \cos \frac \theta 2} = \sqrt{\frac 12 + \frac 12 \sqrt{\frac 12 + \frac 12 \cos \theta}} & \small \color{#3D99F6} \text{Similarly} \\ \cos \frac \theta 8 & = \sqrt{\frac 12 + \frac 12 \sqrt{\frac 12 + \sqrt{\frac 12 + \frac 12 \cos \theta}}} & \small \color{#3D99F6} \text{Multiply both sides by 2} \\ 2 \cos \frac \theta 8 & = \sqrt{2 + \sqrt{2 + \sqrt{2 + \cos \theta}}} & \small \color{#3D99F6} \text{Setting }\theta = \frac \pi 2 \\ 2 \cos \frac \pi {16} & = \sqrt{2 + \sqrt{2 + \sqrt 2}} \end{aligned}$

Therefore, $A+B+C = 2+1+16 = \boxed{16}$ .

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Bonus: We note that $2 \cos \dfrac \pi{2^{n+1}} = \underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots \sqrt{2}}}}}_{\text{Number of 2}=n}$ . Therefore $\displaystyle \lim_{n \to \infty} 2 \cos \dfrac \pi{2^{n+1}} = 2 \cos 0 = 2$ .