Originally Posted by
e^(i*pi) It's the phase difference between cos(x) and sin(x). It can be shown using the addition formula for cos.
$\displaystyle \cos(A-B) = \cos A\cos B + \sin A \sinB$. When $\displaystyle A = \frac{\pi}{2}$ and $\displaystyle b = \theta$ you get:
$\displaystyle \cos \left(\dfrac{\pi}{2} - \theta \right) = \cos \left(\dfrac{\pi}{2}\right) \cos \theta + \sin \left(\dfrac{\pi}{2}\right) \sin \theta$
Since $\displaystyle \cos \left(\dfrac{\pi}{2}\right) = 0 \text{ and } \sin \left(\dfrac{\pi}{2}\right) = 1$ the equation above simplifies to give $\displaystyle \cos \left(\dfrac{\pi}{2} - \theta \right) = \sin \theta$