Suppose that $F$ is defined via the recurrence relation $$F(1)=1, \qquad F(n)=\sum_{k=1}^{n-1}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr)$$ What is $F(N)$? I don't have any idea how to solve this problem. Only one thing that I've noticed is that: $$ 0=\sum_{k=1}^{n}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr).$$
Edit:
From the comment section:
I was trying to rewrite $\sin\left(\frac{\pi}{2^n}\right)$ by the formula for a double argument and I've ended up with $$\sin\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2}\right)}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)}=\frac{1}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)},n\gt 1,$$ but it doesn't seem to help. $F$ is used in another formula. It should be true for most of the functions $$g(x)=\sum_{n=1}^\infty \left(\sin\left(x2^{n-1}\right)\sum_{k=1}^n\left(F(k)g\left(\frac{\pi}{2^{n-k+1}}\right)\right)\right),\;x\in\left(0,\frac{\pi}{2}\right)$$ First four values of $F$ are:
\begin{align*}F(1)&=1\\F(2)&=-\sin\left(\frac{\pi}{4}\right)\\F(3)&=-\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{4}\right)\\F(4)&=-\sin\left(\frac{\pi}{16}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)-\sin^3\left(\frac{\pi}{4}\right)\end{align*}
These terms look like if they created some pattern, but the fifth term which is \begin{align*}F(5)=-\sin\left(\frac{\pi}{32}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{16}\right)-3\sin^2\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)+\sin^4\left(\frac{\pi}{4}\right)\end{align*} meses the pattern up.
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