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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.

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

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

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  • $\begingroup$ Hi, I don't understand where the term $\frac{7\pi}{6}$ is from. $\endgroup$ – Gennaro Arguzzi May 23 '16 at 16:49
  • $\begingroup$ @GennaroArguzzi Note that $\frac{2\pi}{3}+\frac{\pi}{2}=\frac{7\pi}{6}$. We use the first equality. $\endgroup$ – S.C.B. May 23 '16 at 16:50
  • $\begingroup$ @GennaroArguzzi No, we have that $$\cos(\frac{\pi}{2}+\frac{2\pi}{3}-x)=\sin(\frac{2\pi}{3}-x)$$ $\endgroup$ – S.C.B. May 23 '16 at 16:55
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$$ \sin\left(x-\frac{2}{3}\pi\right)=-\sin\left(\frac{2}{3}\pi-x\right)= $$ $$ =-\cos\left(\frac{2}{3}\pi-x-\frac{\pi}{2}\right)=-\cos\left(\frac{4}{6}\pi-x-\frac{3\pi}{6}\right)=-\cos\left(\frac{\pi}{6}-x\right) $$

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

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