Symmetric infinite well potential

Assume we have the following potential :

$$ Vleft(xright)=begin{cases}
0 & -frac{L}{2}leq xleqfrac{L}{2}
infty & else
end{cases} $$

The wave function for a particle in this well potential, is given by :

$ psi_{even}=sqrt{frac{2}{L}}cosleft(frac{npi x}{L}right),thinspacethinspacepsi_{odd}=sqrt{frac{2}{L}}sinleft(frac{npi x}{L}right) $

Where $n$ is even for the sin function, and $n $ is odd for the cos function, as we can see the calculation here

(the last comment)

But I tried to find the odd and even wavefunctions by shifting the wavefunction of a particle in an infinite box (which is not symmetric), and got different result. I’ll be glad if someone can explain the difference. Here’s my calculation:

We know that for a particle in a potential $$ Vleft(xright)=begin{cases}
0 & 0leq xleq L
infty & else
end{cases} $$

The wave function is given by

$ psileft(xright)=sqrt{frac{2}{L}}sinleft(frac{pi n}{L}xright) $

For a symmetric potential we’ll shift the function to the left by $ xto x+frac{L}{2} $ (we can see that for $x=-L/2 $ and for $x=L$ the wave function is $0$ so it preserved).

Now $ psileft(xright)=sqrt{frac{2}{L}}sinleft(frac{pi n}{L}left(x+frac{L}{2}right)right)=sqrt{frac{2}{L}}sinleft(frac{pi n}{L}x+frac{pi n}{2}right) $

If we’ll write for odd $n$ , $ n=2m+1 $ and for even $n$, $n=2m$, we’ll get the solution:

$ psileft(xright)=sqrt{frac{2}{L}}sinleft(frac{pi n}{L}left(x+frac{L}{2}right)right)=sqrt{frac{2}{L}}sinleft(frac{pi n}{L}x+frac{pi n}{2}right)=begin{cases}
psi_{n=2m}=left(-1right)^{m}sqrt{frac{2}{L}}sinleft(frac{pi n}{L}xright)
psi_{n=2m+1}=left(-1right)^{m}sqrt{frac{2}{L}}cosleft(frac{pi n}{L}xright)
end{cases} $

Which is different from the previous result (by a factor of $-1 $ in some of the functions)

What went wrong?

Thanks in advance.