# Taking repeated sin then cos

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?

• It will be the solution to $x=\sin(\cos(x))$. Wolfram alpha has several digits:m.wolframalpha.com/input/… – Brian Rushton Jan 12 '14 at 4:06
• Don't know anything cool about it, though. – Brian Rushton Jan 12 '14 at 4:06
• @BrianRushton. Could you explain for me how you arrived to this simple and smart reformulation of this equation ? Thanks. – Claude Leibovici Jan 12 '14 at 6:01

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.
• Using Newton's method on $f(x)=\sin(\cos(x))-x$, starting with $x_0=0$, it converges very rapidly. $x_1=\sin(1)\approx 0.8415$, $x_2\approx 0.7007$, $x_3\approx 0.6948$. – Glen O Jan 12 '14 at 4:59
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.