Find the minimum value of $n$ such that $\sin^n(c)<\varepsilon$ for some small constant $\varepsilon>0$

Question

Let $c$ be a constant such that $0 <c \le \pi/2$ and $\sin(c) \ne 0$.

Question: What is the minimum value of $n$ such that $\sin^n(c)< \varepsilon$ for some small constant $\varepsilon >0$ ?

For example, let $c=1$ then $\sin(1)= 0.841, \sin^2(1)=0.708, \sin^4(1)=0.501,..., \sin^{20}(1)=0.0316$.

It is clear that $\sin^n(c)$ converges to $0$ as $n$ increases.

Answer 1

The iterated sine (just search for it and you will find, for example, this: http://web.abo.fi/fak/mnf/mate/kurser/dynsyst/Iteration%20of%20sin(x).pdf) satisfies $\sin^{(n)}(x) \approx \sqrt{\frac{3}{n}} $.

Therefore, you want $\sqrt{\frac{3}{n}} < \epsilon $ or $n >\frac{3}{\epsilon^2} $.

Note that $x$ does not really come into this, because, after a few iterations, any dependence on $x$ goes away.

Answer 2

The logarithmic function is strictly increasing, so $\sin^n(c)<\epsilon$ exactly when $\ln(\sin^n(c))<\ln(\epsilon)$, which gives us $$n\ln(\sin(c))<\ln(\epsilon)$$ Now, since $0<c\le\pi/2$, it follows that $0<\sin(c)\le1$. We will consider two cases: 1)Suppose $\sin(c)=1$ $(c=\pi/2).$ Then we have a problem, since $1^n=1$, which will not be less than $\epsilon<1$.

2) Suppose that $\sin(c)<1$. Then $\ln(\sin(c))<0$. Thus, dividing this from the above equation gives us $$n>\frac{\ln(\epsilon)}{\ln(\sin(c))}$$