Prove $\sin(-x) = -\sin(x)$

$$ \sin(x) = \cos(\tfrac\pi2 - x) = \underbrace{\cos(\tfrac\pi2)}_{=0}\cos(-x) - \underbrace{\sin(\tfrac\pi2)}_{=1}\sin(-x) = -\sin(-x) $$

A right-angled triangle with an incline of $x$ has height $\sin x$. A similar triangle with incline $-x$ has (signed) height $\sin(-x)$, and since this is just a reflection of our original triangle in the $x$-axis...

Use the fact that $-x=0-x$ to get $$\sin(-x)=\sin(0-x)=\sin 0\cos x -\cos 0\sin x =0\cos x-1\sin x =-\sin x$$ as required.