Lösung 4.3:4d

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With the formula for double angles and the Pythagorean identity
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<center> [[Image:4_3_4d.gif]] </center>
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<math>\cos ^{2}v+\sin ^{2}v=1</math>, we can express
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<math>\text{cos 2}v\text{ }</math>
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in terms of
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<math>\text{cos }v</math>,
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<math>\begin{align}
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& \text{cos 2}v=\cos ^{2}v-\sin ^{2}v=\cos ^{2}v-\left( 1-\cos ^{2}v \right) \\
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& =2\cos ^{2}v-1=2b^{2}-1 \\
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\end{align}</math>

Version vom 11:48, 29. Sep. 2008

With the formula for double angles and the Pythagorean identity \displaystyle \cos ^{2}v+\sin ^{2}v=1, we can express \displaystyle \text{cos 2}v\text{ } in terms of \displaystyle \text{cos }v,


\displaystyle \begin{align} & \text{cos 2}v=\cos ^{2}v-\sin ^{2}v=\cos ^{2}v-\left( 1-\cos ^{2}v \right) \\ & =2\cos ^{2}v-1=2b^{2}-1 \\ \end{align}

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