From Förberedande kurs i matematik 1

Revision as of 09:58, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.3:8a moved to Solution 4.3:8a: Robot: moved page) ← Previous diff		Current revision (08:16, 10 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that
-	<center> [[Image:4_3_8a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}
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		+	If we then use the Pythagorean identity
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		+	{{Displayed math||<math>\cos^2\!v + \sin^2\!v = 1</math>}}
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		+	and rewrite <math>\cos^2\!v</math> in the denominator as <math>1 - \sin^2\!v</math>, we get what we are looking for on the right-hand side. The whole calculation is
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		+	{{Displayed math||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

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Current revision

We rewrite \displaystyle \tan v on the left-hand side as \displaystyle \frac{\sin v}{\cos v}, so that

\displaystyle \tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}

If we then use the Pythagorean identity

\displaystyle \cos^2\!v + \sin^2\!v = 1

and rewrite \displaystyle \cos^2\!v in the denominator as \displaystyle 1 - \sin^2\!v, we get what we are looking for on the right-hand side. The whole calculation is

\displaystyle \tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}