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Lösung 4.3:3c

Aus Online Mathematik Brückenkurs 1

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With the help of the Pythagorean identity, we can express <math>\cos v</math> in terms of <math>\sin v</math>,
With the help of the Pythagorean identity, we can express <math>\cos v</math> in terms of <math>\sin v</math>,
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{{Displayed math||<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \cos v = \pm\sqrt{1-\sin^2 v}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \cos v = \pm\sqrt{1-\sin^2 v}\,\textrm{.}</math>}}
In addition, we know that the angle <math>v</math> lies between <math>-\pi/2</math>
In addition, we know that the angle <math>v</math> lies between <math>-\pi/2</math>
and <math>\pi/2</math>, i.e. either in the first or fourth quadrant, where angles always have a positive ''x''-coordinate (cosine value); thus, we can conclude that
and <math>\pi/2</math>, i.e. either in the first or fourth quadrant, where angles always have a positive ''x''-coordinate (cosine value); thus, we can conclude that
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{{Displayed math||<math>\cos v = \sqrt{1-\sin^2 v} = \sqrt{1-a^2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\cos v = \sqrt{1-\sin^2 v} = \sqrt{1-a^2}\,\textrm{.}</math>}}

Version vom 08:54, 22. Okt. 2008

With the help of the Pythagorean identity, we can express cosv in terms of sinv,

cos2v+sin2v=1cosv=1−sin2v. 

In addition, we know that the angle v lies between −2 and 2, i.e. either in the first or fourth quadrant, where angles always have a positive x-coordinate (cosine value); thus, we can conclude that

cosv=1−sin2v=1−a2. 
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:3c“