# How can 2 cos &#x2061;<!-- ⁡ --> ( x &#x2212;<!-- - --> &#x03C0;<!-- π --> 2

How can $2\mathrm{cos}\left(x-\frac{\pi }{2}\right)=-2\mathrm{sin}\left(x-\frac{\pi }{2}\right)$
I know that $\mathrm{cos}\left(-x\right)=\mathrm{cos}\left(x\right)$ and that $\mathrm{cos}\left(\frac{\pi }{2}-x\right)=\mathrm{sin}\left(x\right)$
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$2\mathrm{cos}\left(x-\frac{\pi }{2}\right)=-2\mathrm{sin}\left(x-\frac{\pi }{2}\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{tan}\left(x-\frac{\pi }{2}\right)=-1=\mathrm{tan}\left(-\frac{\pi }{4}\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x-\frac{\pi }{2}=n\pi -\frac{\pi }{4}$
where n is any integer
Clearly, the given relationship is not an identity