Lösung 4.2:5c

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If we express the angle
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<center> [[Image:4_2_5c.gif]] </center>
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<math>\text{33}0^{\circ }</math>
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in radians, we obtain
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<math>\text{33}0^{\circ }=\text{33}0^{\circ }\centerdot \frac{\pi }{180^{\circ }}</math>
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radians
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<math>=\frac{11\pi }{6}</math>
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radians
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and from exercise 3.3:1g, we know that
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<math>\cos 330^{\circ }=\cos \frac{11\pi }{6}=\frac{\sqrt{3}}{2}</math>.

Version vom 08:09, 29. Sep. 2008

If we express the angle \displaystyle \text{33}0^{\circ } in radians, we obtain


\displaystyle \text{33}0^{\circ }=\text{33}0^{\circ }\centerdot \frac{\pi }{180^{\circ }} radians \displaystyle =\frac{11\pi }{6} radians

and from exercise 3.3:1g, we know that


\displaystyle \cos 330^{\circ }=\cos \frac{11\pi }{6}=\frac{\sqrt{3}}{2}.

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