Lösung 4.3:4c

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The formula for double angles gives
The formula for double angles gives
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{{Displayed math||<math>\sin 2v=2\sin v\cos v</math>}}
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{{Abgesetzte Formel||<math>\sin 2v=2\sin v\cos v</math>}}
and from exercise b, we have <math>\sin v = \sqrt{1-b^2}\,</math>. Thus,
and from exercise b, we have <math>\sin v = \sqrt{1-b^2}\,</math>. Thus,
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{{Displayed math||<math>\sin 2v = 2b\sqrt{1-b^2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin 2v = 2b\sqrt{1-b^2}\,\textrm{.}</math>}}

Version vom 08:55, 22. Okt. 2008

The formula for double angles gives

\displaystyle \sin 2v=2\sin v\cos v

and from exercise b, we have \displaystyle \sin v = \sqrt{1-b^2}\,. Thus,

\displaystyle \sin 2v = 2b\sqrt{1-b^2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:4c“