Lösung 4.3:4b

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If we once again use the Pythagorean identity we get
If we once again use the Pythagorean identity we get
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{{Displayed math||<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \sin v = \pm\sqrt{1-\cos^2 v}\,\textrm{.}</math>}}
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<math>\cos ^{2}v+\sin ^{2}v=1\quad \Leftrightarrow \quad \sin v=\pm \sqrt{1-\cos ^{2}v}</math>
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Because the angle ''v'' lies between <math>0</math> and <math>\pi</math>, <math>\sin v</math> is positive (an angle in the first and second quadrants has a positive ''y''-coordinate) and therefore
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{{Displayed math||<math>\sin v = +\sqrt{1-\cos^2 v} = \sqrt{1-b^2}\,\textrm{.}</math>}}
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Because the angle v lies between
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<math>0</math>
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and
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<math>\pi </math>,
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<math>\text{sin }v</math>
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is positive (an angle in the first and second quadrants has a positive
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<math>y</math>
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-coordinate) and therefore
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<math>\sin v=+\sqrt{1-\cos ^{2}v}=\sqrt{1-b^{2}}</math>
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Version vom 14:10, 9. Okt. 2008

If we once again use the Pythagorean identity we get

Vorlage:Displayed math

Because the angle v lies between \displaystyle 0 and \displaystyle \pi, \displaystyle \sin v is positive (an angle in the first and second quadrants has a positive y-coordinate) and therefore

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:4b“