Lösung 4.2:4f
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
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If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value, | If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\tan\Bigl(-\frac{5\pi}{3}\Bigr) | \tan\Bigl(-\frac{5\pi}{3}\Bigr) | ||
= \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr) | = \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr) | ||
Version vom 08:52, 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} |
