Proof

Let $x=k\pi$ be a real number by some integer $k\in\mathbb Z$ and the number pi $\pi.$
By representing the real sine by complex function, we have $$\sin\left(\frac x2\right)=\frac{1}{2i}\left(\exp\left(i\frac{x}{2}\right)-\exp\left(-i\frac{x}{2}\right)\right)=\frac{\exp\left(-i\frac x2\right)}{2i}(\exp(ix)-1).$$
Therefore, $\exp(ix)=1$ if and only if $\sin\left(\frac x2\right)=0.$
By zeros of sine, it follows $$\sin\left(\frac x2\right)=0\Longleftrightarrow \frac x2= m\pi\Longleftrightarrow x=2m\pi,\quad m\in\mathbb Z.$$
Therefore, $\exp(ix)=1$ if and only if $k=2m$ is even.
By representing the real cosine by complex function, we have $$\cos\left(\frac x2\right)=\frac{1}{2}\left(\exp\left(i\frac{x}{2}\right)+\exp\left(-i\frac{x}{2}\right)\right)=\frac{\exp\left(-i\frac x2\right)}{2}(\exp(ix)+1).$$
Therefore, $\exp(ix)=-1$ if and only if $\cos\left(\frac x2\right)=0.$
By zeros of cosine, it follows $$\cos\left(\frac x2\right)=0\Longleftrightarrow \frac x2= \left(m+\frac 12\right)\pi=x=(2m+1)\pi,\quad m\in\mathbb Z.$$
Therefore, $\exp(ix)=-1$ if and only if $k=2m+1$ is odd.
∎

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