Lösung 4.2:4a

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It can be a little difficult to draw the angle
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<center> [[Image:4_2_4a.gif]] </center>
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<math>\frac{11\pi }{6}</math>
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straight onto a unit circle, but if we rewrite
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<math>\frac{11\pi }{6}</math>
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as
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<math>\frac{11\pi }{6}=\frac{6\pi +3\pi +2\pi }{6}=\pi +\frac{\pi }{2}+\frac{\pi }{3}</math>
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we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.
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We also note that this angle corresponds to exactly the same point on the unit circle as the angle
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<math>-\frac{\pi }{6}</math>, and because we calculated
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<math>\cos \left( -\frac{\pi }{6} \right)</math>
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in exercise f, we have that
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<math>\cos \frac{11\pi }{6}=\cos \left( -\frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}</math>
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[[Image:4_2_4_a.gif|center]]
[[Image:4_2_4_a.gif|center]]

Version vom 12:52, 28. Sep. 2008

It can be a little difficult to draw the angle \displaystyle \frac{11\pi }{6} straight onto a unit circle, but if we rewrite \displaystyle \frac{11\pi }{6}

as

\displaystyle \frac{11\pi }{6}=\frac{6\pi +3\pi +2\pi }{6}=\pi +\frac{\pi }{2}+\frac{\pi }{3}


we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.

We also note that this angle corresponds to exactly the same point on the unit circle as the angle \displaystyle -\frac{\pi }{6}, and because we calculated \displaystyle \cos \left( -\frac{\pi }{6} \right) in exercise f, we have that


\displaystyle \cos \frac{11\pi }{6}=\cos \left( -\frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}


Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:4a“