Sum of Sine and Cosine to Higher Powers Is Constant (Recursion, Proofs) I have observed this empirically, but I have no idea how to prove it or if it has been proven before.  If this has been proven before, in any form, either more or less generic, please point me to such a proof.  If not, can you prove it?
Given: 
$$ \begin{align}
& n\in \mathbb{\{1, 2, 3, \cdots\}} \\
& c_n=\left( \frac{1}{2}\right)^{2^n-1}\\
& f_1(x)=\sin(x)\\
& f_n(x)=f_{n-1}^2(x)-c_n\\
& g_1(x)=\cos(x)\\
& g_n(x)=g_{n-1}^2\left(x-\frac{\pi}{2^n}\right)-c_n \end{align}$$
Prove that:
$$f_n^2(x)+g_n^2(x)=\left( \frac{1}{2} \right)^{2^n-2}$$
Here are some graphs for reference: https://www.desmos.com/calculator/01pshoohrl

The trick here is that they're all versions of the Pythagorean identity. Your iteration $f_n(x)=f_{n-1}^2(x)-c_n$ and $g_n(x)=g_{n-1}^2(x-\frac{\pi}{2^n})-c_n$ is a double-angle formula. We have $f_2(x)=\sin^2 x-\frac12 = -\frac12\cos(2x)$, $g_2(x)=\cos^2(x-\frac{\pi}{4})-\frac12=\frac12\cos(2(x-\frac{\pi}{4}))=\frac12\cos(2x-\frac{\pi}{2})$
$=\frac12\sin(2x)$. By a similar process, $f_3(x)=\frac14\cos^2(2x)-\frac18=\frac18\cos(4x)$ and $g_3(x)=\frac14\sin^2(2x-2\frac{\pi}{8})-\frac18=-\frac18\cos(4x-\frac{\pi}{2})=-\frac18\sin(4x)$.
It stabilizes from there, and $f_n(x)=2^{1-2^{n-1}}\cos\left(2^{n-1}x\right)$, $g_n(x)=-2^{1-2^{n-1}}\sin\left(2^{n-1}x\right)$ for all $n\ge 3$.
To prove the result, then, we prove these formulas inductively. We do the special case calculations for $n=1$, $n=2$, and $n=3$, then do the inductive step to go from $n-1$ to $n$ for $n\ge 4$. In all of these cases, we're using the double-angle formulas $\sin^2 t-\frac12=-\frac12\cos(2t)$ and $\cos^2 t-\frac12=\frac12\cos(2t)$ and the translation formula $\cos(t-\frac{\pi}{2})=\sin t$.
Once we have the formulas, the result about the sum of squares is immediate:
$$f_n^2(x)+g_n^2(x)=\left(2^{1-2^{n-1}}\right)^2\left(\cos^2(2^{n-1}x)+\sin^2(2^{n-1}x)\right)=2^{2-2^n}$$
(and the special cases for small $n$)
