Lösung 4.4:1f

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We can translate the equation
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<center> [[Image:4_4_1f.gif]] </center>
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<math>\sin v=-\frac{1}{2}</math>
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to the problem of finding those angles in the unit circle which have a y-coordinate of
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<math>-\frac{1}{2}</math>. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied
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<math>\sin v=+\frac{1}{2}</math>, then the situation is the same, except that the angles now lie under, rather than above, the
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<math>x</math>
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-axis, due to reflectional symmetry.
[[Image:4_4_1_f.gif|center]]
[[Image:4_4_1_f.gif|center]]
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Angle
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<math>2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>
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Angle
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<math>\pi +\frac{\pi }{6}=\frac{7\pi }{6}</math>
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The two angles which satisfy
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<math>\sin v=-\frac{1}{2}</math>
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lie in the third and fourth quadrants and are
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<math>v=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>
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and
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<math>v=\pi +\frac{\pi }{6}=\frac{7\pi }{6}</math>

Version vom 12:47, 30. Sep. 2008

We can translate the equation \displaystyle \sin v=-\frac{1}{2} to the problem of finding those angles in the unit circle which have a y-coordinate of \displaystyle -\frac{1}{2}. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied \displaystyle \sin v=+\frac{1}{2}, then the situation is the same, except that the angles now lie under, rather than above, the \displaystyle x -axis, due to reflectional symmetry.


Angle \displaystyle 2\pi -\frac{\pi }{6}=\frac{11\pi }{6} Angle \displaystyle \pi +\frac{\pi }{6}=\frac{7\pi }{6}


The two angles which satisfy \displaystyle \sin v=-\frac{1}{2} lie in the third and fourth quadrants and are \displaystyle v=2\pi -\frac{\pi }{6}=\frac{11\pi }{6} and \displaystyle v=\pi +\frac{\pi }{6}=\frac{7\pi }{6}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.4:1f“