Showing that $c_n=\cos\frac\pi{2^n}$ satisfies the recurrence $c_{n+1} = \sqrt{\frac{1+c_n}2}$ From the identity $\cos(\alpha) = \sqrt{\frac{1+\cos(2\alpha)}{2}}$ with $0 \leq \alpha \leq \frac{\pi}{2}$, show that the sequence $c_n = \cos(\frac{\pi}{2^n})$ satisfies the recurrence relation $$c_{n+1} = \sqrt{\frac{1+c_n}{2}}$$

What I did was:
Say that $\alpha_n = \frac{\pi}{2^n}$. We first must prove $0 \leq \alpha_n \leq \frac{\pi}{2}$. By induction:
Base case is trivial: $0 \leq \frac{\pi}{2^1} \leq \frac{\pi}{2}$.
For $p(n) \Rightarrow p(n+1)$:
$0 \leq \frac{\pi}{2^n} \leq \frac{\pi}{2} \iff 0 \leq \frac{\pi}{2^n}\frac{1}{2} \leq \frac{\pi}{2}\frac{1}{2} \iff 0 \leq \frac{\pi}{2^{n+1}} \leq \frac{\pi}{4} &lt; \frac{\pi}{2}$
So it is true that $\{a_n\}^\infty_{n=1} \in [0, \frac{\pi}{2}]$ and we can write $c_n = \cos(a_n)$.
Therefore:
$$c_1 = \cos(\alpha_1) = \cos(\frac{\pi}{2}) = 0$$
$$c_n = \cos(\alpha) \iff 1+c_n = 1+\cos(\alpha)\iff \frac{1+c_n}{2} = \frac{1+\cos(\alpha)}{2}$$
$$\iff \sqrt{\frac{1+c_n}{2}} = \sqrt{\frac{1+\cos(\alpha)}{2}}$$
But $\sqrt{\frac{1+c_n}{2}}$ would be the following term for $c_n$, meaning $c_{n+1} = \sqrt{\frac{1+c_n}{2}}$.
This was my atempt to prove it but I can't tell if it actually proves the recurrence. How could I prove recurrence relations in sequences like this?
You're doing much more work than you have to. As @Surb notes, you just need to verify $\cos\tfrac{\pi}{2^{n+1}}=\sqrt{\tfrac{1+\cos\tfrac{\pi}{2^n}}{2}}$, which is the case $\alpha=\tfrac{\pi}{2^{n+1}}$.
Let $\alpha = \frac{\pi}{2^{n+1}}$
$c_{n+1} = \cos\frac{\pi}{2^{n+1}} = \cos\alpha$ and $c_n = \cos\frac{\pi}{2^n} =  \cos\frac{2\pi}{2^{n+1}} = \cos(2\alpha)$
Now, $\cos(\alpha) = \sqrt{\dfrac{1+\cos(2\alpha)}{2}} \Rightarrow c_{n+1} = \sqrt{\dfrac{1+c_n}{2}}$