Lösung 4.2:4b

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We start by subtracting
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We start by subtracting <math>2\pi</math> from <math>11\pi/3</math>, so that we get an angle between <math>0</math> and <math>2\pi </math>. This doesn't change the cosine value
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<math>2\pi </math>
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from
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<math>\frac{11\pi }{3}</math>, so that we get an angle between
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<math>o</math>
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and
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<math>2\pi </math>. This doesn't change the cosine value
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{{Displayed math||<math>\cos\frac{11\pi}{3} = \cos\Bigl(\frac{11\pi}{3}-2\pi\Bigr) = \cos\frac{5\pi}{3}\,\textrm{.}</math>}}
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<math>\cos \frac{11\pi }{3}=\cos \left( \frac{11\pi }{3}-2\pi \right)=\cos \frac{5\pi }{3}</math>
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Then, by rewriting <math>5\pi/3</math> as a sum of <math>\pi</math>- and <math>\pi/2</math>-terms,
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{{Displayed math||<math>\frac{5\pi}{3} = \frac{3\pi +\dfrac{3}{2}\pi +\dfrac{1}{2}\pi}{3} = \pi + \frac{\pi}{2} + \frac{\pi}{6}</math>}}
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Then, by rewriting
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we see that <math>5\pi/3</math> is an angle in the fourth quadrant which makes an angle <math>\pi/6</math> with the negative ''y''-axis.
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<math>\frac{5\pi }{3}</math>
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as a sum of
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<math>\pi </math>
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<math>\frac{\pi }{2}</math>
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-terms
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[[Image:4_2_4b1.gif||center]]
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<math>\frac{5\pi }{3}=\frac{3\pi +\frac{3}{2}\pi +\frac{1}{2}\pi }{3}=\pi +\frac{\pi }{2}+\frac{\pi }{6}</math>
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With the help of an auxiliary triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle
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<math>5\pi/3\,</math>.
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we see that
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{| width="100%"
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<math>\frac{5\pi }{3}</math>
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|width="50%" align="center"|[[Image:4_2_4_b2.gif]]
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is an angle in the fourth quadrant which makes an angle
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|width="50%" align="left"|<math>\begin{align}\text{opposite} &= 1\cdot\sin\frac{\pi}{6} = \frac{1}{2}\\[5pt] \text{adjacent} &= 1\cdot\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\end{align}</math>
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<math>\frac{\pi }{6}</math>
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|}
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with the negative
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<math>y</math>
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-axis.
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The point has coordinates <math>(1/2,-\sqrt{3}/2)</math> and
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[[Image:4_2_4b1.gif]]
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{{Displayed math||<math>\cos \frac{11\pi}{3} = \cos\frac{5\pi}{3} = \frac{1}{2}\,\textrm{.}</math>}}
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With the help of a triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle
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<math>\frac{5\pi }{3}</math> .
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[[Image:4_2_4_b2.gif]]
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The point has coordinates
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<math>\left( \frac{1}{2} \right.,\left. -\frac{\sqrt{3}}{2} \right)</math>
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and
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<math>\cos \frac{11\pi }{3}=\cos \frac{5\pi }{3}=\frac{1}{2}</math>.
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Version vom 08:37, 9. Okt. 2008

We start by subtracting \displaystyle 2\pi from \displaystyle 11\pi/3, so that we get an angle between \displaystyle 0 and \displaystyle 2\pi . This doesn't change the cosine value

Vorlage:Displayed math

Then, by rewriting \displaystyle 5\pi/3 as a sum of \displaystyle \pi- and \displaystyle \pi/2-terms,

Vorlage:Displayed math

we see that \displaystyle 5\pi/3 is an angle in the fourth quadrant which makes an angle \displaystyle \pi/6 with the negative y-axis.

With the help of an auxiliary triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle \displaystyle 5\pi/3\,.

Image:4_2_4_b2.gif \displaystyle \begin{align}\text{opposite} &= 1\cdot\sin\frac{\pi}{6} = \frac{1}{2}\\[5pt] \text{adjacent} &= 1\cdot\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\end{align}

The point has coordinates \displaystyle (1/2,-\sqrt{3}/2) and

Vorlage:Displayed math

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