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

Dec 2015
939
122
Earth
Express \(\displaystyle \sin(1/n)\) in terms of \(\displaystyle \sin(n) \; \) , where \(\displaystyle n\in \mathbb{N} \).
 
Mar 2015
182
68
Universe 2.71828i3.14159
Use magic.
And where did you get this problem from?
Generalize the problem for \(\displaystyle z\), where \(\displaystyle z \in \mathbb{C}\).
 
  • Like
Reactions: idontknow
Dec 2015
939
122
Earth
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 ?
 
Mar 2015
182
68
Universe 2.71828i3.14159
No, your answer should be in terms of \(\displaystyle \sin(n)\), not in terms of \(\displaystyle \sin(arcofnoah(n))\)
 

mathman

Forum Staff
May 2007
6,895
760
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
Sep 2015
2,773
1,552
USA
I'm thinking the MacLaurin series should be the first step
 

v8archie

Math Team
Dec 2013
7,709
2,677
Colombia
The question sounds like a Fourier Series question, but since $\sin\frac1n$ isn't periodic, there's a problem.
 
  • Like
Reactions: idontknow

romsek

Math Team
Sep 2015
2,773
1,552
USA
yeah wait... do you know if this problem has a solution? Or are you just slinging random ... stuff... out there?
 
  • Like
Reactions: idontknow