Lösung 4.2:3d

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{{NAVCONTENT_START}}
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In order to get an angle between
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<center> [[Image:4_2_3d.gif]] </center>
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<math>0</math>
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{{NAVCONTENT_STOP}}
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and
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<math>\text{2}\pi </math>, we subtract
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<math>\text{2}\pi </math>
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from
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<math>{7\pi }/{2}\;</math>
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, which also leaves the cosine value unchanged
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<math>\cos \frac{7\pi }{2}=\cos \left( \frac{7\pi }{2}-2\pi \right)=\cos \frac{3\pi }{2}</math>
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When we draw a line which makes an angle
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<math>{3\pi }/{2}\;</math>
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with the positive
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<math>x</math>
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-axis, we get the negative
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<math>y</math>
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-axis and we see that this line cuts the unit circle at the point
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<math>\left( 0 \right.,\left. -1 \right)</math>. The
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<math>x</math>
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-coordinate of the intersection point is thus
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<math>0</math>
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and hence
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<math>\cos {7\pi }/{2}\;=\cos {3\pi }/{2}\;=0</math>
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[[Image:4_2_3_d.gif|center]]
[[Image:4_2_3_d.gif|center]]

Version vom 12:06, 28. Sep. 2008

In order to get an angle between \displaystyle 0 and \displaystyle \text{2}\pi , we subtract \displaystyle \text{2}\pi from \displaystyle {7\pi }/{2}\; , which also leaves the cosine value unchanged


\displaystyle \cos \frac{7\pi }{2}=\cos \left( \frac{7\pi }{2}-2\pi \right)=\cos \frac{3\pi }{2}


When we draw a line which makes an angle \displaystyle {3\pi }/{2}\; with the positive \displaystyle x -axis, we get the negative \displaystyle y -axis and we see that this line cuts the unit circle at the point \displaystyle \left( 0 \right.,\left. -1 \right). The \displaystyle x -coordinate of the intersection point is thus \displaystyle 0 and hence \displaystyle \cos {7\pi }/{2}\;=\cos {3\pi }/{2}\;=0


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