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

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
$$n>\frac{\ln(\epsilon)}{\ln(\sin(c))}$$
