Lösung 4.2:4e

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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If we write the angle <math>\frac{7\pi}{6}</math> as
If we write the angle <math>\frac{7\pi}{6}</math> as
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{{Displayed math||<math>\frac{7\pi}{6} = \frac{6\pi+\pi}{6} = \pi + \frac{\pi }{6}</math>}}
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{{Abgesetzte Formel||<math>\frac{7\pi}{6} = \frac{6\pi+\pi}{6} = \pi + \frac{\pi }{6}</math>}}
we see that the angle <math>7\pi/6</math> on the unit circle is in the third quadrant and makes an angle <math>\pi/6</math> with the negative ''x''-axis.
we see that the angle <math>7\pi/6</math> on the unit circle is in the third quadrant and makes an angle <math>\pi/6</math> with the negative ''x''-axis.
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Geometrically, <math>\tan (7\pi/6)</math> is defined as the slope of the line having an angle <math>7\pi/6</math> and, because this line has the same slope as the line having angle <math>\pi/6</math>, we have that
Geometrically, <math>\tan (7\pi/6)</math> is defined as the slope of the line having an angle <math>7\pi/6</math> and, because this line has the same slope as the line having angle <math>\pi/6</math>, we have that
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{{Displayed math||<math>\tan\frac{7\pi}{6} = \tan\frac{\pi}{6} = \frac{\sin\dfrac{\pi }{6}}{\cos\dfrac{\pi }{6}} = \frac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\tan\frac{7\pi}{6} = \tan\frac{\pi}{6} = \frac{\sin\dfrac{\pi }{6}}{\cos\dfrac{\pi }{6}} = \frac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\,\textrm{.}</math>}}
[[Image:4_2_4_e2.gif|center]]
[[Image:4_2_4_e2.gif|center]]

Version vom 08:52, 22. Okt. 2008

If we write the angle \displaystyle \frac{7\pi}{6} as

\displaystyle \frac{7\pi}{6} = \frac{6\pi+\pi}{6} = \pi + \frac{\pi }{6}

we see that the angle \displaystyle 7\pi/6 on the unit circle is in the third quadrant and makes an angle \displaystyle \pi/6 with the negative x-axis.

Geometrically, \displaystyle \tan (7\pi/6) is defined as the slope of the line having an angle \displaystyle 7\pi/6 and, because this line has the same slope as the line having angle \displaystyle \pi/6, we have that

\displaystyle \tan\frac{7\pi}{6} = \tan\frac{\pi}{6} = \frac{\sin\dfrac{\pi }{6}}{\cos\dfrac{\pi }{6}} = \frac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:4e“