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Stuck on homework problem (not this), if I can prove as a lemma that the sequence $$\sin(\sin(\sin\cdots(\sin1)\cdots) \rightarrow 0 $$ then I'm done. It's monotonic and decreasing and bounded by 0 and 1 respectively, so it converges, though is it truly $0$ ?

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    $\begingroup$ Hint: what solutions are there to the equation x = sin(x) $\endgroup$ – John Apr 3 '14 at 20:19
  • $\begingroup$ yup, now I get it... wonder if that's what they expect of us. $\endgroup$ – user7610 Apr 3 '14 at 20:27
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    $\begingroup$ See math.stackexchange.com/questions/45283/… $\endgroup$ – Rustyn Apr 3 '14 at 22:15
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Hint: From what you stated, you've proven that it converges to some limit, call it $L$. You know that $0 \leq L \leq 1$. Do you see an identity that $L$ satisfies with respect to sine?

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    $\begingroup$ sin(L)=L as they're both limits of the sequence? and then L=0. Wow. $\endgroup$ – user7610 Apr 3 '14 at 20:27
  • $\begingroup$ I treated this problem as a fixed point and tried to prove it via banach but it's seems like $\sin(x)$ is not a contraction map on $[0,1]$. Or not? $\endgroup$ – rlartiga Apr 3 '14 at 21:14
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This is a recursive function. Start with a definition: $$\sin(x)=x$$ This is already a dead giveaway, the only place where this holds true is at $x=0$. This is the answer, it's this simple. The $\sin(1)$ in the middle is a little misleading.

As you can see here, it's slowly getting down to $0$ as you approach infinity. Also note that the test one ($\sin 1337$) isn't much different from the original because this recursive function will hold true for whatever is inside at the end.

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