Lösung 4.3:8c

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One could write
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<center> [[Image:4_3_8c-1(2).gif]] </center>
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<math>\tan \frac{u}{2}</math>
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as a quotient involving sine and cosine, and then continue with the formula for half-angles,
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<math>\tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...</math>
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but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.
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We write
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<math>u</math>
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as
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<math>2\left( \frac{u}{2} \right)</math>
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and use the formula for double angles (so as to end up with a right-hand side which has
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<math>\frac{u}{2}</math>
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as its argument)
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<math>\frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}</math>
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Writing the
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<math>\text{1}</math>
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in the denominator as
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<math>\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}</math>
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using the Pythagorean identity,
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<math>\begin{align}
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& \frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}=\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}} \\
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& =\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{2\cos ^{2}\frac{u}{2}}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\tan \frac{u}{2} \\
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\end{align}</math>

Version vom 11:08, 30. Sep. 2008

One could write \displaystyle \tan \frac{u}{2} as a quotient involving sine and cosine, and then continue with the formula for half-angles,


\displaystyle \tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...


but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.

We write \displaystyle u as \displaystyle 2\left( \frac{u}{2} \right) and use the formula for double angles (so as to end up with a right-hand side which has \displaystyle \frac{u}{2} as its argument)


\displaystyle \frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}


Writing the \displaystyle \text{1} in the denominator as \displaystyle \cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2} using the Pythagorean identity,


\displaystyle \begin{align} & \frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}=\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}} \\ & =\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{2\cos ^{2}\frac{u}{2}}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\tan \frac{u}{2} \\ \end{align}

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