Prove that $\sin(x)=0 \iff x=k \pi$, where $k$ is a natural number?

Prove that $\sin(x)=0 \iff x=k \pi$, where $k$ is a natural number.

Definitions:

$\pi:=2\cdot \inf\{x>0:\cos(x)=0\},$
$\sin(z):=\frac{1}{2i}(e^{iz}-e^{-iz}),$
$\cos(z):=\frac{1}{2}(e^{iz}+e^{-iz}),$ $\exp(z):=\sum_{k=0}^{\infty} \frac{z^{k}}{k!}.$

I could show that $\sin(k \pi)=0$, but I have trouble showing that $k\pi$ are the only roots.

Solution 1:

From the definitions, you can see that $\cos(-x) = \cos (x)$, $\cos (0) =1$ and $\sin^2(z)+\cos^2(z) =1$.

Now, you can check using the binomial theorem that $e^{z+w}=e^ze^w$, and this can be used to prove the formulas

$$\cos(z+w)=\cos(w) \cos(z) -\sin(z)\sin(w)$$

As has been already done in this post.

Therefore, $\cos (\pi) = 2\cos^2(\pi/2)-1 = -1$, $\cos(2 \pi) = 2\cos^2 (\pi)-1 = 1$, so $\sin (2 \pi )=0$, and so

$$\cos(x+ 2 \pi) = \cos(x) \cos (2 \pi) - \sin (x) \sin (2 \pi) =\cos(x)$$

$\cos$ is periodic with period $2 \pi$.

We know that $\sin^2(\pi/2)=1$, so let $\alpha = \sin(\pi/2)$, which a priori could be $1$ or $-1$. Then,

$$\cos(x+\pi/2) = -\alpha \sin (x) = -\cos (\pi/2-x)$$

This shows that $\cos$ is symmetric with change of sign around the line $x = \pi/2$. Therefore, since $\cos$ does not have any zeros in the interval $[0, \pi/2)$ (by your definition of $\pi$, it does not have zeros in the interval $(\pi/2, \pi]$, so $\pi/2$ is its inly zero in $[0,\pi]$. But $\cos(-x) = \cos(x)$, so

in $[-\pi,\pi]$, the only zeros of $\cos$ are $\pm \pi/2$.

The combination of the two highlighted facts we have proven shows that $\cos(x) = 0$ if and only if $x = \left(k+\frac{1}{2}\right)\pi$ for some $k \in \mathbb Z$.

Finally, we go back to the equality $\cos(x+\pi/2) = -\alpha \sin (x)$ which tells us that $\sin (x) =0$ if and only if $\cos(x + \pi/2)=0$, and this proves the claim.