# Convergence of nested sin and cos

Let's define the following: $A_1 = x, A_j = \cos(\sin A_{j - 1})$.

I notice that $\forall x \in \mathbb{R}$, $A_\infty \approx 0.768169157$

Is there an exact value for $A_\infty$?

• I remember noticing that as a high-schooler. I don't think there's any interesting closed-form expression for it, but maybe someone will chime in. – Brian Tung Mar 12 '18 at 16:59

Yes, for all $A_1\in \mathbb{R}$ the recurrence $(A_n)_n$ converges to $A_{\infty}$ which is the unique fixed point of the contraction mapping $f(x)=\cos(\sin(x))$. This means that $A_{\infty}$ is the only solution of the equation $$x = \cos(\sin (x)).$$ I'm afraid that the real number $A_{\infty}\approx 0.768169156736795977$ has not a "closed form".
P.S. The function $f$ is a contraction over $\mathbb{R}$ because for any $x\in\mathbb{R}$, $$|f'(x)|=|-\sin(\sin(x))\cos(x)|\leq \sin(1)<1.$$