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lemon14 · Dec 6, 2010, 11:54 AM

1. $\sin \frac{\pi}{3} - i \cos \frac{\pi}{3}$
$\varphi = \arctan \frac{\cos \frac{\pi}{3}}{\sin \frac{\pi}{3}} + \pi = \arccot \frac{\pi}{3} +\pi = \arctan \[tan$\frac{\pi}{2}+\frac{\pi}{3}$\]+\pi = -\frac{\pi}{2}-\frac{\pi}{3}+\pi = \frac{11\pi}{6}$

2. [math]\sin\varphi - i \cos[\varphi, \varphi \epsilon \(-\frac{\pi}{2},\pi)/math]
$-\arctan \frac{\cos \varphi}{\sin \varphi} + 2\pi = -\arctan \cot \varphi+2\pi = \varphi - \frac{\pi}{2}$

I have a problem in understanding how these were calculated. In the first case, I understood that $\varphi$ has been calculated for $\frac{\pi}{2},\pi$ , but in the second case for $0,\frac{\pi}{2}$ , but why? There is the same sign.

lemon14 · Dec 6, 2010, 11:55 AM

2. $\sin\varphi - i \cos[\varphi, \varphi \epsilon \(-\frac{\pi}{2},\pi)$