Lösung 4.2:3f

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The point on the unit circle which corresponds to the angle
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<center> [[Image:4_2_3f-1(2).gif]] </center>
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<math>-{\pi }/{6}\;</math>
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lies in the fourth quadrant.
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<center> [[Image:4_2_3f-2(2).gif]] </center>
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[[Image:4_2_3_f1.gif]]
[[Image:4_2_3_f1.gif]]
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As usual,
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<math>\cos \left( -{\pi }/{6}\; \right)</math>
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gives the
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<math>x</math>
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-coordinate of the point of intersection between the angle's line and the unit circle. In order to determine this point, we introduce an auxiliary triangle in the fourth quadrant.
[[Image:4_2_3_f2.gif]]
[[Image:4_2_3_f2.gif]]
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We can determine the edges in this triangle by simple trigonometry and then translate these over to the point's coordinates.
[[Image:4_2_3_f3.gif]]
[[Image:4_2_3_f3.gif]]
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The coordinates of the point of intersection are
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<math>\left( \frac{\sqrt{3}}{2} \right.,\left. -\frac{1}{2} \right)</math>
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and in particular
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<math>\cos \left( -{\pi }/{6}\; \right)=\frac{\sqrt{3}}{2}</math>.

Version vom 12:35, 28. Sep. 2008

The point on the unit circle which corresponds to the angle \displaystyle -{\pi }/{6}\; lies in the fourth quadrant.

Image:4_2_3_f1.gif

As usual, \displaystyle \cos \left( -{\pi }/{6}\; \right) gives the \displaystyle x -coordinate of the point of intersection between the angle's line and the unit circle. In order to determine this point, we introduce an auxiliary triangle in the fourth quadrant.

Image:4_2_3_f2.gif

We can determine the edges in this triangle by simple trigonometry and then translate these over to the point's coordinates.

Image:4_2_3_f3.gif

The coordinates of the point of intersection are \displaystyle \left( \frac{\sqrt{3}}{2} \right.,\left. -\frac{1}{2} \right) and in particular \displaystyle \cos \left( -{\pi }/{6}\; \right)=\frac{\sqrt{3}}{2}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:3f“