# Is $\sin(n^k) ≠ (\sin n)^k$ in general?

Is it true that $\sin(n^k) ≠ (\sin n)^k$ for any positive integers $n$ and integers $k ≠ 1$?

What if $n > 0, k ≠ 1$ are rational?

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@Chandru: :) [ ](http://.) – kennytm Aug 13 '10 at 13:26

If $\sin(n^k)=\sin(n)^k$, then $\frac{e^{in^k}-e^{-in^k}}{2i}=(\frac{e^{in}-e^{-in}}{2i})^k$, so that $e^i$ would be algebraic.
But that directly contradicts the Lindemann-Weierstrass theorem because $i$ is algebraic.