Lösung 4.2:4e

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If we write the angle
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<center> [[Image:4_2_4e.gif]] </center>
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<math>\frac{7\pi }{6}</math>
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{{NAVCONTENT_STOP}}
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as
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<math>\frac{7\pi }{6}=\frac{6\pi +\pi }{6}=\pi +\frac{\pi }{6}</math>
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we see that the angle
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<math>\frac{7\pi }{6}</math>
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on a unit circle is in the third quadrant and makes an angle
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<math>\frac{\pi }{6}</math>
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with the negative
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<math>x</math>
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-axis.
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[[Image:4_2_4_e1.gif|center]]
[[Image:4_2_4_e1.gif|center]]
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Geometrically,
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<math>\tan \frac{7\pi }{6}</math>
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is defined as the gradient of the line having an angle
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<math>\frac{7\pi }{6}</math>
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and, because this line has the same slope as the line having angle
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<math>\frac{\pi }{6}</math>, we have that
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<math>\tan \frac{7\pi }{6}=\tan \frac{\pi }{6}=\frac{\sin \frac{\pi }{6}}{\cos \frac{\pi }{6}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}</math>
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[[Image:4_2_4_e2.gif|center]]
[[Image:4_2_4_e2.gif|center]]

Version vom 13:23, 28. Sep. 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 \frac{7\pi }{6} on a unit circle is in the third quadrant and makes an angle \displaystyle \frac{\pi }{6} with the negative \displaystyle x -axis.

Geometrically, \displaystyle \tan \frac{7\pi }{6} is defined as the gradient of the line having an angle \displaystyle \frac{7\pi }{6} and, because this line has the same slope as the line having angle \displaystyle \frac{\pi }{6}, we have that

\displaystyle \tan \frac{7\pi }{6}=\tan \frac{\pi }{6}=\frac{\sin \frac{\pi }{6}}{\cos \frac{\pi }{6}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}


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