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>, | ||
| - | {{ | + | {{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 | ||
| - | {{ | + | {{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 \displaystyle \cos v in terms of \displaystyle \sin v,
| \displaystyle \cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \cos v = \pm\sqrt{1-\sin^2 v}\,\textrm{.} |
In addition, we know that the angle \displaystyle v lies between \displaystyle -\pi/2 and \displaystyle \pi/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
| \displaystyle \cos v = \sqrt{1-\sin^2 v} = \sqrt{1-a^2}\,\textrm{.} |
