Lösung 4.2:5b

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If we draw the angle
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If we draw the angle <math>225^{\circ} = 180^{\circ} + 45^{\circ}</math> on a unit circle, we see that it makes an angle of <math>45^{\circ}</math> with the negative ''x''-axis.
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<math>\text{225}^{\circ }\text{ }=\text{ 18}0^{\circ }\text{ }+\text{ 45}^{\circ }</math>
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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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This means that <math>\tan 225^{\circ}</math>, which is the slope of the line that makes an angle of <math>45^{\circ}</math> with the positive ''x''-axis, equals <math>\tan 45^{\circ}</math>, because the line which makes an angle of <math>45^{\circ}</math> has the same slope,
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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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{{Displayed math||<math>\tan 225^{\circ} = \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1\,\textrm{.}</math>}}
[[Image:4_2_5_b2.gif|center]]
[[Image:4_2_5_b2.gif|center]]

Version vom 11:08, 9. Okt. 2008

If we draw the angle \displaystyle 225^{\circ} = 180^{\circ} + 45^{\circ} on a unit circle, we see that it makes an angle of \displaystyle 45^{\circ} with the negative x-axis.

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

Vorlage:Displayed math

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