# sin(1/n) in terms of sin(n)

#### idontknow

Express $$\displaystyle \sin(1/n)$$ in terms of $$\displaystyle \sin(n) \;$$ , where $$\displaystyle n\in \mathbb{N}$$.

#### tahirimanov19

Use magic.
And where did you get this problem from?
Generalize the problem for $$\displaystyle z$$, where $$\displaystyle z \in \mathbb{C}$$.

idontknow

#### idontknow

Let $$\displaystyle \sin(n)=t$$ , $$\displaystyle 1/n=1/\arcsin(t)$$ , both sides -sin.
$$\displaystyle \sin(1/n)=\sin(1/\arcsin(t))$$.

Can we accept this solution ?

#### tahirimanov19

No, your answer should be in terms of $$\displaystyle \sin(n)$$, not in terms of $$\displaystyle \sin(arcofnoah(n))$$

#### mathman

Forum Staff
Highly unlikely $\lim_{n\to \infty} \sin(1/n)=0$ while $\sin(n)$ does not converge to anything as $n \to \infty$.

#### romsek

Math Team
I'm thinking the MacLaurin series should be the first step

#### v8archie

Math Team
The question sounds like a Fourier Series question, but since $\sin\frac1n$ isn't periodic, there's a problem.

idontknow

#### romsek

Math Team
yeah wait... do you know if this problem has a solution? Or are you just slinging random ... stuff... out there?

idontknow

#### idontknow

Yes it has solution.