Lösung 4.2:5d

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K (Lösning 4.2:5d moved to Solution 4.2:5d: Robot: moved page)
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By subtracting
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<center> [[Image:4_2_5d.gif]] </center>
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<math>360^{\circ }</math>
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from
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<math>\text{495}^{\circ }</math>, we do not change the value of the tangent:
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<math>\tan \text{495}^{\circ }=\tan \left( \text{495}^{\circ }-360^{\circ } \right)=\tan \text{135}^{\circ }</math>
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We know from exercise a that
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<math>\cos 135^{\circ }=-\frac{1}{\sqrt{2}}</math>
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and
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<math>\sin 135^{\circ }=\frac{1}{\sqrt{2}}</math>, which gives
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<math>\tan 135^{\circ }=\frac{\sin 135^{\circ }}{\cos 135^{\circ }}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1</math>

Version vom 08:16, 29. Sep. 2008

By subtracting \displaystyle 360^{\circ } from \displaystyle \text{495}^{\circ }, we do not change the value of the tangent:


\displaystyle \tan \text{495}^{\circ }=\tan \left( \text{495}^{\circ }-360^{\circ } \right)=\tan \text{135}^{\circ }

We know from exercise a that \displaystyle \cos 135^{\circ }=-\frac{1}{\sqrt{2}} and \displaystyle \sin 135^{\circ }=\frac{1}{\sqrt{2}}, which gives


\displaystyle \tan 135^{\circ }=\frac{\sin 135^{\circ }}{\cos 135^{\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_4.2:5d“