Lösung 4.2:4f
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Version vom 14:57, 22. Okt. 2008
If we add \displaystyle 2\pi to \displaystyle -5\pi/3\,, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle \displaystyle -5\pi/3 and consequently has the same tangent value,
| \displaystyle \begin{align}
\tan\Bigl(-\frac{5\pi}{3}\Bigr) = \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr) = \tan\frac{\pi}{3} = \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}\,\textrm{.} \end{align} |
