Making $\sin(2k\pi x)$ an odd function about $x=\frac{1}{4}$

Question

I want to ask whether k should be even or odd in order to make the function $$f(x)=\sin(2k\pi x)$$ to be an odd function about $x=\frac{1}{4}$.
I think that if it is an odd function, then, $$f(x+\frac{1}{4})=-f(x-\frac{1}{4})$$. I sub it in; but it turns out I get to: $$\sin(2k\pi x)\cos(\frac{1}{2}k\pi)x=0.$$
Then, I don’t know how to proceed.
Thank you so much.

Answer 1

Let $y=x-\frac 1 4$. If $k$ is odd then $f(x)=\sin (2k\pi y+k\pi /2)=\sin (k\pi /2) \cos (2k\pi y)$. This is an even function of $y$. On the other hand if $k$ is even then $f(x)=\sin(2k\pi y) \cos (k\pi /2)$ which is an odd function of $y$.

Answer 2

$k$ should be even

Because if $\sin(2\pi kx)$ is odd about $x=\frac{1}{4}$, then you're asking for what $k$ does $\sin\Big(2\pi k(\frac{1}{4})\Big)=0$

I.e;

$$\sin\Big(\frac{\pi k}{2}\Big)=0$$

This is because $\sin(\theta)$ is odd about $\theta$ only when $\sin(\theta)=0$. Therefore

$$\sin\Big(\frac{\pi k}{2}\Big)=0\implies\frac{\pi k}{2}=\pi n~~~~~\forall n\in\mathbb{Z}$$

$$\implies k=2n~~~~~\forall n\in\mathbb{Z}$$

So $k$ can be any even integer.