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
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We rewrite
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<math>\text{tan }v</math>
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on the left-hand side as
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<math>\frac{\sin v}{\cos v}</math>, so that
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<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}</math>
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{{Displayed math||<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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<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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and rewrite
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<math>\text{cos}^{\text{2}}v</math>
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in the denominator as
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<math>\text{1}-\text{sin}^{\text{2}}v\text{ }</math>, we get what we are looking for on the right-hand side. The whole calculation is
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<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}</math>
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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>}}

Version vom 08:16, 10. Okt. 2008

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

Vorlage:Displayed math

If we then use the Pythagorean identity

Vorlage:Displayed math

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

Vorlage:Displayed math

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