What is $ \sin(x)+\sin(x−π)+\sin(x+π) $? So I have this trig question:

$ \sin(x)+\sin(x−π)+\sin(x+π) = $ _____

The answer is $- \sin(x)$
I can't figure out how to solve it. 
Any help?
 A: As shown in some other answers, this is very simple if you know that:
$$\sin(x-\pi)=-\sin x \quad\mbox{and}\quad \sin(x+\pi)=-\sin x$$
If you don't know these formulas or you have a hard time understanding why they are true, you should spend some time to carefully study the unit circle and how symmetry there leads to these simple relations.
The image below should help you understand why $\sin(x+\pi)=-\sin x$.

Then note that by "adding a full cirle", the same holds for the angle $x-\pi$:
$$\sin(x-\pi)=\sin(x-\pi\color{blue}{+2\pi})=\sin(x+\pi)=-\sin x$$
A: $$\sin(x)+\color{green}{\sin(x-\pi)}+\color{red}{\sin(x+\pi)}$$ $$=\sin(x)+\color{green}{\sin(x)\cos(-\pi)+\cos(x)\sin(-\pi)}+\color{red}{\sin(x)\cos(\pi)+\cos(x)\sin(\pi)}$$
$$=\sin(x)\color{green}{-\sin(x)}\color{red}{-\sin(x)}=-\sin(x)$$
using the formula for $\sin(x+\theta)$ and the facts that $\cos(\pm\pi)=-1$ and $\sin(\pm\pi)=0$
A: You probably know, that $$ \sin(x−\pi) = -\sin(x).$$
Also $$\sin(x+\pi) = \sin(x-\pi + 2\pi) = \sin(x-\pi)$$ so your given expression reduces to $$\sin x - \sin x - \sin x$$
A: Note that $\sin(\pi -x)=\sin x$ and $\sin(\pi+x)=-\sin x$, using which we get:
$$\begin{aligned}\lambda&=\sin x+\sin(x-\pi)+\sin(x+\pi)\\&= \sin x-\sin(\pi -x)+\sin(\pi+x)\\&=\sin x-\sin x-\sin x=-\sin x\end{aligned}$$
$$\sin(\pi -x)=\sin \pi \cos x-\sin x\cos\pi=+\sin x \\ \sin(\pi+x)=\sin\pi\cos x+\sin x\cos \pi =-\sin x$$
