Lösung 4.2:4c

Aus Online Mathematik Brückenkurs 1

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In exercise e, we studied the angle
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<center> [[Image:4_2_4c.gif]] </center>
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<math>\frac{3\pi }{4}</math>
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and found that
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<math>\cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}}</math>
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and
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<math>\sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}</math>
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Because
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<math>\text{tan }x</math>
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is defined as
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<math>\frac{\sin x}{\cos x}</math>, we get immediately that
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<math>\tan \frac{3\pi }{4}=\frac{\sin \frac{3\pi }{4}}{\cos \frac{3\pi }{4}}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1</math>

Version vom 13:11, 28. Sep. 2008

In exercise e, we studied the angle \displaystyle \frac{3\pi }{4} and found that

\displaystyle \cos \frac{3\pi }{4}=-\frac{1}{\sqrt{2}} and \displaystyle \sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}


Because \displaystyle \text{tan }x is defined as \displaystyle \frac{\sin x}{\cos x}, we get immediately that


\displaystyle \tan \frac{3\pi }{4}=\frac{\sin \frac{3\pi }{4}}{\cos \frac{3\pi }{4}}=\frac{\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}}=-1

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