Lösung 4.3:3d

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{{NAVCONTENT_START}}
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The expression for the angle
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<center> [[Image:4_3_3d.gif]] </center>
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<math>{\pi }/{2}\;-v</math>
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{{NAVCONTENT_STOP}}
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differs from
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<math>{\pi }/{2}\;</math>
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by as much as
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<math>-v\text{ }</math>
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differs from
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<math>0</math>. This means that
 +
<math>{\pi }/{2}\;</math>
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makes the same angle with the positive
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<math>y</math>
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-axis as
 +
<math>-v\text{ }</math>
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makes with the positive
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<math>x</math>
 +
-axis.
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 +
 
[[Image:4_3_3_d.gif|center]]
[[Image:4_3_3_d.gif|center]]
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 +
Angle
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<math>v</math>
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angle
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<math>\pi -v</math>
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 +
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Therefore, the angle
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<math>{\pi }/{2}\;-v</math>
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has a
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<math>y</math>
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-coordinate which is equal to the
 +
<math>x</math>
 +
-coordinate for the angle
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<math>v</math>, i.e.
 +
 +
 +
<math>\sin \left( {\pi }/{2}\;-v \right)=\cos v</math>
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 +
 +
and from exercise c, we know that
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<math>\cos v=\sqrt{1-a^{2}}</math>
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 +
 +
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<math>\sin \left( \frac{\pi }{2}-v \right)=\sqrt{1-a^{2}}</math>

Version vom 11:11, 29. Sep. 2008

The expression for the angle \displaystyle {\pi }/{2}\;-v differs from \displaystyle {\pi }/{2}\; by as much as \displaystyle -v\text{ } differs from \displaystyle 0. This means that \displaystyle {\pi }/{2}\; makes the same angle with the positive \displaystyle y -axis as \displaystyle -v\text{ } makes with the positive \displaystyle x -axis.


Angle \displaystyle v angle \displaystyle \pi -v


Therefore, the angle \displaystyle {\pi }/{2}\;-v has a \displaystyle y -coordinate which is equal to the \displaystyle x -coordinate for the angle \displaystyle v, i.e.


\displaystyle \sin \left( {\pi }/{2}\;-v \right)=\cos v


and from exercise c, we know that \displaystyle \cos v=\sqrt{1-a^{2}}


\displaystyle \sin \left( \frac{\pi }{2}-v \right)=\sqrt{1-a^{2}}

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