Lösung 4.3:4b
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Version vom 15:04, 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{.} |
