FindSequenceFunction applied to basic trig sequence

I an not too familiar with FindSequenceFunction.

Is there a trick to make Mathematica produce the obvious sequence function for this sequence?

Example 1

ClearAll[y,x];
op={-y''[x],DirichletCondition[y[x]==0, True]};
eigf=Last@DEigensystem[op,y[x],{x,0,1},8] It is clear that the eigenfunctions are of the form $\sin\left( n \pi x\right)$ for $n=1,2,\dots$ and I thought that using FindSequenceFunction would be the way to show this, so I typed

FindSequenceFunction[eigf,n]

But it gives It works well for things like:

Example 2

ClearAll[y,x,L0];
op={-y''[x],DirichletCondition[y[x]==0,True]};
eig=DEigenvalues[op,y[x],{x,0,L0},8] FindSequenceFunction[eig,n] Question Is there a better function or way to change this to produce the desired result for example 1? I looked at options for FindSequenceFunction. Why did Matghematica do a good job for example 2 and not example 1?

ClearAll[y, x];
op = {-y''[x], DirichletCondition[y[x] == 0, True]};
eigf = Last@DEigensystem[op, y[x], {x, 0, 1}, 8]

(*  {Sin[π x], Sin[2 π x], Sin[3 π x], Sin[4 π x], Sin[5 π x],
Sin[6 π x], Sin[7 π x], Sin[8 π x]}  *)

Use FullSimplify

sf = FindSequenceFunction[eigf, n] // FullSimplify

(*  1/2 I ((E^(-I π x))^n - (E^(I π x))^n)  *)

For integer values of n

sf // FullSimplify[#, Element[n, Integers]] &

(*  Sin[n π x]  *)