Equivalence between trigonometric functions

Question

could you help me to understand why:

$\sin\left(x-\frac{2\pi}{3}\right)=-\cos\left(\frac{\pi}{6}-x\right)$

?

Thank you for your help.

Answer 1

Use $$\sin\left(x\right)=\cos\left(\frac{\pi}{2}-x\right)$$ $$\cos (\pi+x)=-\cos x$$ The result follows.

One can use $$\sin\left(x-\frac{2\pi}{3}\right)=\cos\left(\frac{7\pi}{6}-x\right)=-\cos\left(\frac{\pi}{6}-x\right)$$

Answer 2

$$ \sin\left(x-\frac{2}{3}\pi\right)=-\sin\left(\frac{2}{3}\pi-x\right)= $$ $$ =-\cos\left(\frac{2}{3}\pi-x-\frac{\pi}{2}\right)=-\cos\left(\frac{4}{6}\pi-x-\frac{3\pi}{6}\right)=-\cos\left(\frac{\pi}{6}-x\right) $$

where I only used the facts that $\sin\left(-x\right)=-\sin\left(x\right)$ and $\sin\left(x\right)=\cos\left(x-\frac{\pi}{2}\right)$