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When you keep taking alternating sin and cos of any number as follows:

$$\sin(\cos(\sin(\cos(\sin(\cos...(N))))...)$$

it seems to converge at about 0.69. Is there any way to find the exact value it converges at?

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    $\begingroup$ It will be the solution to $x=\sin(\cos(x))$. Wolfram alpha has several digits:m.wolframalpha.com/input/… $\endgroup$ – Brian Rushton Jan 12 '14 at 4:06
  • $\begingroup$ Don't know anything cool about it, though. $\endgroup$ – Brian Rushton Jan 12 '14 at 4:06
  • $\begingroup$ @BrianRushton. Could you explain for me how you arrived to this simple and smart reformulation of this equation ? Thanks. $\endgroup$ – Claude Leibovici Jan 12 '14 at 6:01
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The exact value may not have a nice closed form. If you take $x^*$ to be the point that this converges to, then we have \begin{equation} \sin(\cos(x^*)) = x^* \end{equation} which naturally gives us \begin{equation} \cos(x^*) = \arcsin(x^*). \end{equation} Looking at their intersection using Wolfram Alpha here gives us the numerical approximation $x^* = 0.694819690730788...$ and I don't happen to recognize this as any familiar fraction of $\pi$ or some such expression at the moment.

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    $\begingroup$ Using Newton's method on $f(x)=\sin(\cos(x))-x$, starting with $x_0=0$, it converges very rapidly. $x_1=\sin(1)\approx 0.8415$, $x_2\approx 0.7007$, $x_3\approx 0.6948$. $\endgroup$ – Glen O Jan 12 '14 at 4:59
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If taking sin and cos repeatedly has a limit, call it x. Then $\sin(\cos(\sin(\cos...)))=x$, and so, taking sin and cos of both sides, $\sin(\cos(\sin(\cos(\sin(\cos...)))))=\sin(\cos(x))$. Since the wo left sides. The quations are equal, so are the right sides.

This can be made rigorous if neccessary.

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