Lösung 4.2:4f

Aus Online Mathematik Brückenkurs 1

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If we add
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<center> [[Image:4_2_4f.gif]] </center>
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<math>2\pi </math>
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to
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<math>-\frac{5\pi }{3}</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle
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<math>-\frac{5\pi }{3}</math>
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and consequently has the same tangent value:
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<math>\begin{align}
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& \tan \left( -\frac{5\pi }{3} \right)=\tan \left( -\frac{5\pi }{3}+2\pi \right)=\tan \frac{\pi }{3} \\
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& =\frac{\sin \frac{\pi }{3}}{\cos \frac{\pi }{3}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \\
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\end{align}</math>

Version vom 13:27, 28. Sep. 2008

If we add \displaystyle 2\pi to \displaystyle -\frac{5\pi }{3}, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle \displaystyle -\frac{5\pi }{3} and consequently has the same tangent value:


\displaystyle \begin{align} & \tan \left( -\frac{5\pi }{3} \right)=\tan \left( -\frac{5\pi }{3}+2\pi \right)=\tan \frac{\pi }{3} \\ & =\frac{\sin \frac{\pi }{3}}{\cos \frac{\pi }{3}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \\ \end{align}

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