Lösung 4.3:4c

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The formula for double angles gives
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<center> [[Image:4_3_4c.gif]] </center>
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<math>\sin 2v=2\sin v\cos v</math>
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and from exercise b, we have
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<math>\sin v=\sqrt{1-b^{2}}</math>
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Thus,
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<math>\sin 2v=2b\sqrt{1-b^{2}}</math>

Version vom 11:44, 29. Sep. 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}}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:4c“