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How do I prove that $$\frac{1}{45}<\sin^{2020}\left(\frac{\pi}4\right)<\frac{2}{45},$$ where $\sin^n$ denotes the composition of the sine function with itself $n$ times. For example, $$\sin^3(x) = \sin(\sin(\sin(x))).$$

Are there any relations between $45$ and $\sin x?$ or is there some way to calculate an approximation？

Xander Henderson
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pudding eggs
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    Related: https://math.stackexchange.com/q/1609995. – Martin R Apr 30 '21 at 13:11
  • Where does the question come from? How did you come up with $1/45$ and $2/45$ as lower/upper bound? – Martin R Apr 30 '21 at 13:19
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    @puddingeggs The notation $f^{(n)}$ nearly always denotes the $n$-th derivative of $f$. On the other hand, the notation $f^n = f\circ f\circ f \circ \dotsb \circ f$ (the $n$-fold composition of $f$ with itself) is pretty common. I have edited your question to use this more standard notation, and to improve the grammar a bit. However, it would still be good if you could expand on the question a bit. As Martin R has suggested, the bounds you have given seen a bit odd---where did they come from? What theory do you expect should be relevant? – Xander Henderson Apr 30 '21 at 15:44
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    Perhaps, $\sin_n(\frac{\pi}{4})$ or $\sin^{\circ n}(\frac{\pi}{4})$ is OK. $\sin^n x$ looks like n-th power. – River Li May 01 '21 at 00:38

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Let $x_n = \sin _n \left( {\frac{\pi }{4}} \right)$ where the subscript $n$ indicates the number of times $\sin$ is iterated. I leave it as an exercise to show that $$ 1 + \frac{{x^2 }}{3} \leq \left( {\frac{{x }}{{\sin x }}} \right)^2\leq 1 + \frac{{x^2 }}{2} $$ for $|x|\leq 1$. With these inequalities $$ \frac{1}{{x_{n + 1}^2 }} = \frac{1}{{(\sin x_n )^2 }} = \frac{1}{{x_n^2 }}\left( {\frac{{x_n }}{{\sin x_n }}} \right)^2 \geq \frac{1}{{x_n^2 }} + \frac{1}{3} $$ and $$ \frac{1}{{x_{n + 1}^2 }} = \frac{1}{{(\sin x_n )^2 }} = \frac{1}{{x_n^2 }}\left( {\frac{{x_n }}{{\sin x_n }}} \right)^2 \leq \frac{1}{{x_n^2 }} + \frac{1}{2}. $$ Thus $$ \frac{{n - 1}}{3} \leq \sum\limits_{k = 1}^{n - 1} {\left( {\frac{1}{{x_{k + 1}^2 }} - \frac{1}{{x_k^2 }}} \right)} = \frac{1}{{x_n^2 }} - \frac{1}{{x_1^2 }} = \frac{1}{{x_n^2 }} - 2 $$ and $$ \frac{{n - 1}}{2} \geq \sum\limits_{k = 1}^{n - 1} {\left( {\frac{1}{{x_{k + 1}^2 }} - \frac{1}{{x_k^2 }}} \right)} = \frac{1}{{x_n^2 }} - \frac{1}{{x_1^2 }} = \frac{1}{{x_n^2 }} - 2. $$ Therefore, $$ \sqrt {\frac{2}{{n + 3}}} \leq x_n \leq \sqrt {\frac{3}{{n + 5}}} . $$ In particular, $$ \frac{1}{{45}} < \sqrt {\frac{2}{{2023}}} \leq x_{2020} \le \sqrt {\frac{3}{{2025}}} < \frac{2}{{45}}. $$

Gary
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