From Förberedande kurs i matematik 1

Revision as of 07:14, 21 August 2008 (edit) Tekbot (Talk | contribs) m (Robot: Automated text replacement (-[[Bild: +[[Image:)) ← Previous diff		Current revision (14:10, 9 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we once again use the Pythagorean identity we get
-	<center> [[Image:4_3_4b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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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		+	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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Current revision

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{.}