Lösung 4.3:8c

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One could write <math>\tan\frac{u}{2}</math> as a quotient involving sine and cosine, and then continue with the formula for half-angles,
One could write <math>\tan\frac{u}{2}</math> as a quotient involving sine and cosine, and then continue with the formula for half-angles,
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{{Displayed math||<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}
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{{Abgesetzte Formel||<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}
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.
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.
Zeile 7: Zeile 7:
We write <math>u</math> as <math>2\cdot(u/2)</math>and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument),
We write <math>u</math> as <math>2\cdot(u/2)</math>and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument),
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{{Displayed math||<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}
Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity,
Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}}
\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}}
&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt]
&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt]

Version vom 08:57, 22. Okt. 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\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots

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\cdot(u/2)and use the formula for double angles (so as to end up with a right-hand side which has \displaystyle u/2 as its argument),

\displaystyle \frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}

Writing the 1 in the denominator as \displaystyle \cos^2(u/2) + \sin^2(u/2) using the Pythagorean identity,

\displaystyle \begin{align}

\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}} &= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt] &= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{2\cos^2\dfrac{u}{2}}\\[5pt] &= \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}}\\[8pt] &= \tan\frac{u}{2}\,\textrm{.} \end{align}

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