Lösung 4.2:5a

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{{NAVCONTENT_START}}
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Because
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<center> [[Image:4_2_5a.gif]] </center>
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<math>\text{135}^{\circ }\text{ }=\text{ 9}0^{\circ }\text{ }+\text{45}^{\circ }</math>,
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{{NAVCONTENT_STOP}}
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<math>\text{135}^{\circ }\text{ }</math>
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is an angle in the second quadrant which makes an angle of
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<math>\text{45}^{\circ }</math>
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with the positive
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<math>y</math>
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-axis.
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[[Image:4_2_5_a1.gif|center]]
[[Image:4_2_5_a1.gif|center]]
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We can determine the point on the unit circle which corresponds to
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<math>\text{135}^{\circ }\text{ }</math>
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by introducing an auxiliary triangle and calculating its edges using trigonometry.
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[[Image:4_2_5_a2.gif|center]]
[[Image:4_2_5_a2.gif|center]]
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opposite<math>=1\centerdot \sin \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>
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adjacent
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<math>=1\centerdot \cos \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>
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The coordinates of the point are
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<math>\left( -\frac{1}{\sqrt{2}} \right.,\left. \frac{1}{\sqrt{2}} \right)</math>
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and this shows that
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<math>\text{cos135}^{\circ }=-\frac{1}{\sqrt{2}}</math>.

Version vom 07:56, 29. Sep. 2008

Because \displaystyle \text{135}^{\circ }\text{ }=\text{ 9}0^{\circ }\text{ }+\text{45}^{\circ }, \displaystyle \text{135}^{\circ }\text{ } is an angle in the second quadrant which makes an angle of \displaystyle \text{45}^{\circ } with the positive \displaystyle y -axis.


We can determine the point on the unit circle which corresponds to \displaystyle \text{135}^{\circ }\text{ } by introducing an auxiliary triangle and calculating its edges using trigonometry.


opposite\displaystyle =1\centerdot \sin \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}

adjacent \displaystyle =1\centerdot \cos \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}


The coordinates of the point are \displaystyle \left( -\frac{1}{\sqrt{2}} \right.,\left. \frac{1}{\sqrt{2}} \right) and this shows that \displaystyle \text{cos135}^{\circ }=-\frac{1}{\sqrt{2}}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:5a“