How do I prove that $\frac{1}{45}<\sin^{2020}(\frac{\pi}4)<\frac{2}{45}?$ 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？

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,
$$
\frac{1}{{45}} < \sqrt {\frac{2}{{2023}}}  \leq x_{2020}  \le \sqrt {\frac{3}{{2025}}}  < \frac{2}{{45}}.
$$
