Lösung 4.2:5b

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K (Lösning 4.2:5b moved to Solution 4.2:5b: Robot: moved page)
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{{NAVCONTENT_START}}
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If we draw the angle
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<center> [[Image:4_2_5b.gif]] </center>
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<math>\text{225}^{\circ }\text{ }=\text{ 18}0^{\circ }\text{ }+\text{ 45}^{\circ }</math>
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{{NAVCONTENT_STOP}}
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on a unit circle, we see that it makes an angle of
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<math>\text{45}^{\circ }</math>
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with the negative
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<math>x</math>
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-axis.
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[[Image:4_2_5_b1.gif|center]]
[[Image:4_2_5_b1.gif|center]]
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This means that
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<math>\text{tan 225}^{\circ }</math>, which is the gradient of the line that 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>x</math>
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-axis, equals
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<math>\text{tan 225}^{\circ }</math>, because the line which makes an angle of
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<math>\text{45}^{\circ }</math>
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has the same slope:
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<math>\tan 225^{\circ }\text{ }=\tan \text{45}^{\circ }=\frac{\sin \text{45}^{\circ }}{\cos \text{45}^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1</math>
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[[Image:4_2_5_b2.gif|center]]
[[Image:4_2_5_b2.gif|center]]

Version vom 08:04, 29. Sep. 2008

If we draw the angle \displaystyle \text{225}^{\circ }\text{ }=\text{ 18}0^{\circ }\text{ }+\text{ 45}^{\circ } on a unit circle, we see that it makes an angle of \displaystyle \text{45}^{\circ } with the negative \displaystyle x -axis.

This means that \displaystyle \text{tan 225}^{\circ }, which is the gradient of the line that makes an angle of \displaystyle \text{45}^{\circ } with the positive \displaystyle x -axis, equals \displaystyle \text{tan 225}^{\circ }, because the line which makes an angle of \displaystyle \text{45}^{\circ } has the same slope:


\displaystyle \tan 225^{\circ }\text{ }=\tan \text{45}^{\circ }=\frac{\sin \text{45}^{\circ }}{\cos \text{45}^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1


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