Lösung 4.3:6b

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We draw an angle
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We draw an angle <math>v</math> in the unit circle, and the fact that <math>\sin v = 3/10</math> means that its ''y''-coordinate equals <math>3/10</math>.
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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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means that its
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<math>y</math>
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-coordinate equals
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<math>\frac{3}{10}</math>.
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[[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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With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of 1 and a vertical side of length 3/10.
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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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We can determine the triangle's remaining side by using the Pythagorean theorem,
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<math>a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}</math>
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{{Displayed math||<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}
which gives that
which gives that
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{{Displayed math||<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}
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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 ''x''-coordinate is <math>-a</math>, i.e. we have
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This means that the angle's
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{{Displayed math||<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}
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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
and thus
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{{Displayed math||<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}
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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>
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Version vom 13:00, 10. Okt. 2008

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

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

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

Vorlage:Displayed math

which gives that

Vorlage:Displayed math

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

Vorlage:Displayed math

and thus

Vorlage:Displayed math

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