How can $2\cos(x-\frac{\pi}2) = -2\sin(x-\frac{\pi}2)$ How can $2\cos(x-\dfrac{\pi}2) = -2\sin(x-\dfrac{\pi}2)$
I know that  $\cos(-x) = \cos(x)$ and that $\cos(\dfrac{\pi}2-x) = \sin(x)$
From these two formulas I can get 
$1.$ $2\cos(x-\dfrac{\pi}2) = 2\cos(\dfrac{\pi}2-x)$ 
$2.$ $2\cos(\dfrac{\pi}2-x) = 2 \sin(x)$ 
$3.$ $ 2\sin(x) = -2\sin(-x)$
$4.$ How can I get $-2\sin(-x) = -2\sin(x-\dfrac{\pi}2)$ 
It was part of calculation of limit: 
$lim_{x->(\dfrac{\pi}{2})} \dfrac{1-e^{2cosx}}{2cosx} \dfrac{2cos(x-\dfrac{\pi}{2})}{sin(4(x-\dfrac{\pi}{x}))}$
and from there in the next step we assumed the identity I mentioned above

$$2\cos\left(x-\dfrac\pi2\right)=-2\sin\left(x-\dfrac\pi2\right)$$
$$\implies\tan\left(x-\dfrac\pi2\right)=-1=\tan\left(-\dfrac\pi4\right)$$
$$\implies x-\dfrac\pi2=n\pi-\dfrac\pi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
