Lösung 4.2:4a

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It can be a little difficult to draw the angle <math>11\pi/6</math> straight onto a unit circle, but if we rewrite <math>11\pi/6</math> as
It can be a little difficult to draw the angle <math>11\pi/6</math> straight onto a unit circle, but if we rewrite <math>11\pi/6</math> as
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{{Displayed math||<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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{{Abgesetzte Formel||<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}
we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.
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 <math>-\pi/6</math>, and because we calculated <math>\cos (-\pi/6)</math> in exercise 4.2:3f, we have that
We also note that this angle corresponds to exactly the same point on the unit circle as the angle <math>-\pi/6</math>, and because we calculated <math>\cos (-\pi/6)</math> in exercise 4.2:3f, we have that
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{{Displayed math||<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}
[[Image:4_2_4_a.gif|center]]
[[Image:4_2_4_a.gif|center]]

Version vom 08:51, 22. Okt. 2008

It can be a little difficult to draw the angle \displaystyle 11\pi/6 straight onto a unit circle, but if we rewrite \displaystyle 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 -\pi/6, and because we calculated \displaystyle \cos (-\pi/6) in exercise 4.2:3f, we have that

\displaystyle \cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.2:4a“