Lösung 4.3:3e

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{{NAVCONTENT_START}}
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The angle
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<center> [[Image:4_3_3e.gif]] </center>
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<math>\frac{\pi }{2}+v</math>
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{{NAVCONTENT_STOP}}
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makes the same angle with the positive
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<math>y</math>
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-axis as the angle
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<math>v</math>
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makes with the positive
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<math>x</math>
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-axis, and hence we see that the
 +
<math>x</math>
 +
-coordinate for
 +
<math>\frac{\pi }{2}+v</math>
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is equal to the
 +
<math>y</math>
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-coordinate for
 +
<math>v</math>, but with a change of sign, i.e.
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 +
 
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<math>\cos \left( \frac{\pi }{2}+v \right)=-\sin v=-a</math>
[[Image:4_3_3_e-1.gif|center]][[Image:4_3_3_e-2.gif|center]]
[[Image:4_3_3_e-1.gif|center]][[Image:4_3_3_e-2.gif|center]]
 +
 +
 +
angle
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<math>v</math>
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angle
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<math>\frac{\pi }{2}+v</math>

Version vom 09:12, 10. Okt. 2008

The angle \displaystyle \frac{\pi }{2}+v makes the same angle with the positive \displaystyle y -axis as the angle \displaystyle v makes with the positive \displaystyle x -axis, and hence we see that the \displaystyle x -coordinate for \displaystyle \frac{\pi }{2}+v is equal to the \displaystyle y -coordinate for \displaystyle v, but with a change of sign, i.e.


\displaystyle \cos \left( \frac{\pi }{2}+v \right)=-\sin v=-a


angle \displaystyle v angle \displaystyle \frac{\pi }{2}+v

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