Lösung 4.3:8b
Aus Online Mathematik Brückenkurs 1
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Version vom 15:08, 22. Okt. 2008
Because \displaystyle \tan v = \frac{\sin v}{\cos v}, the left-hand side can be written using \displaystyle \cos v as the common denominator,
| \displaystyle \frac{1}{\cos v} - \tan v = \frac{1}{\cos v} - \frac{\sin v}{\cos v} = \frac{1-\sin v}{\cos v}\,\textrm{.} |
Now, we observe that if we multiply top and bottom with \displaystyle 1+\sin v, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give \displaystyle 1-\sin^2\!v = \cos ^2\!v\,, using the difference of two squares,
| \displaystyle \begin{align}
\frac{1-\sin v}{\cos v} &= \frac{1-\sin v}{\cos v}\cdot\frac{1+\sin v}{1+\sin v}\\[5pt] &= \frac{1-\sin^2\!v}{\cos v\,(1+\sin v)}\\[5pt] &= \frac{\cos^2\!v}{\cos v\,(1+\sin v)}\,\textrm{.} \end{align} |
Eliminating \displaystyle \cos v then gives the answer,
| \displaystyle \frac{\cos^2\!v}{\cos v\,(1+\sin v)} = \frac{\cos v}{1+\sin v}\,\textrm{.} |
