Lösung 4.3:8a

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We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that
We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that
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{{Displayed math||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}
If we then use the Pythagorean identity
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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{{Abgesetzte Formel||<math>\cos^2\!v + \sin^2\!v = 1</math>}}
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
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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{{Abgesetzte Formel||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

Version vom 08:56, 22. Okt. 2008

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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:8a“