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