Lösung 4.3:8c
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Version vom 15:09, 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} |
