Is $\sin(\omega t+\pi)=-\sin\omega t$ always true no matter what $\omega$ is?

Question

$$\sin(\omega t+\pi)=\sin\omega t$$

I found this in my textbook. But I don't think this property holds for all $\omega$. $t$ is in radians. And $\omega$ in radians/second.

For example, it's true that $\sin(t+\pi)=-\sin(t)$. You flip one over the $x$-axis, and you get the other. But $\sin(2t+\pi)=\sin(2t)$. Because $\sin(2t)$ has a period of $\pi$, so if you translate it by $\pi$, you are still gonna get $\sin(2t)$.

Answer 1

You are wrong: let $u=2t$; then$$\sin(2t+\pi)=\sin(u+\pi)=-\sin(u)=-\sin(2t).$$

You are right when you assert that $\sin(2t+\pi)$ is periodic with period $\pi$, but what you extract from that is that$$\sin\bigl(2(t-\pi)+\pi\bigr)=2\sin(2t+\pi);$$in other words,$$\sin(2t-\pi)=\sin(2t+\pi).$$

Answer 2

Basic trigonometry says that: $$\sin(\alpha + \pi) = -\sin\alpha$$

You can replace $\alpha$ with whatever you want.