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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{{Abgesetzte Formel||<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \sin v = \pm\sqrt{1-\cos^2 v}\,\textrm{.}</math>}}
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
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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{{Abgesetzte Formel||<math>\sin v = +\sqrt{1-\cos^2 v} = \sqrt{1-b^2}\,\textrm{.}</math>}}

Version vom 08:55, 22. Okt. 2008

If we once again use the Pythagorean identity we get

\displaystyle \cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \sin v = \pm\sqrt{1-\cos^2 v}\,\textrm{.}

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

\displaystyle \sin v = +\sqrt{1-\cos^2 v} = \sqrt{1-b^2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:4b“