Lösung 4.3:6b

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{{NAVCONTENT_START}}
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We draw an angle
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<center> [[Image:4_3_6b-1(2).gif]] </center>
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<math>v</math>
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in the unit circle, and the fact that
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<math>\text{sin }v\text{ }=\frac{3}{10}</math>
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<center> [[Image:4_3_6b-2(2).gif]] </center>
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means that its
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{{NAVCONTENT_STOP}}
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<math>y</math>
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-coordinate equals
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<math>\frac{3}{10}</math>.
[[Image:4_3_6_b1.gif|center]]
[[Image:4_3_6_b1.gif|center]]
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With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of
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<math>\text{1}</math>
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and a vertical side of length
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<math>\frac{3}{10}</math>.
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[[Image:4_3_6_b2.gif|center]]
[[Image:4_3_6_b2.gif|center]]
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We can determine the triangle's remaining side by using Pythagoras' theorem,
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<math>a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}</math>
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which gives that
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<math>a=\sqrt{1-\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}</math>
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This means that the angle's
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<math>x</math>
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-coordinate is
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<math>-a</math>, i.e. we have
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<math>\cos v=-\frac{\sqrt{91}}{10}</math>
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and thus
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<math>\tan v=\frac{\sin v}{\cos v}=\frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}</math>

Version vom 09:15, 10. Okt. 2008

We draw an angle \displaystyle v in the unit circle, and the fact that \displaystyle \text{sin }v\text{ }=\frac{3}{10} means that its \displaystyle y -coordinate equals \displaystyle \frac{3}{10}.

With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of \displaystyle \text{1} and a vertical side of length \displaystyle \frac{3}{10}.


We can determine the triangle's remaining side by using Pythagoras' theorem,


\displaystyle a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}


which gives that


\displaystyle a=\sqrt{1-\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}

This means that the angle's \displaystyle x -coordinate is \displaystyle -a, i.e. we have


\displaystyle \cos v=-\frac{\sqrt{91}}{10}

and thus


\displaystyle \tan v=\frac{\sin v}{\cos v}=\frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:6b“