Lösung 4.4:5c
Aus Online Mathematik Brückenkurs 1
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For a fixed value of ''u'', an equality of the form | For a fixed value of ''u'', an equality of the form | ||
- | {{ | + | {{Abgesetzte Formel||<math>\cos u=\cos v</math>}} |
is satisfied by two angles ''v'' in the unit circle, | is satisfied by two angles ''v'' in the unit circle, | ||
- | {{ | + | {{Abgesetzte Formel||<math>v=u\qquad\text{and}\qquad v=-u\,\textrm{.}</math>}} |
[[Image:4_4_5_c.gif|center]] | [[Image:4_4_5_c.gif|center]] | ||
Zeile 11: | Zeile 11: | ||
This means that all angles ''v'' which satisfy the equality are | This means that all angles ''v'' which satisfy the equality are | ||
- | {{ | + | {{Abgesetzte Formel||<math>v=u+2n\pi\qquad\text{and}\qquad v=-u+2n\pi\,,</math>}} |
where ''n'' is an arbitrary integer. | where ''n'' is an arbitrary integer. | ||
Zeile 17: | Zeile 17: | ||
Therefore, the equation | Therefore, the equation | ||
- | {{ | + | {{Abgesetzte Formel||<math>\cos 5x=\cos (x+\pi/5)</math>}} |
has the solutions | has the solutions | ||
- | {{ | + | {{Abgesetzte Formel||<math>\left\{\begin{align} 5x&=x+\frac{\pi}{5}+2n\pi\quad\text{or}\\[5pt] 5x &= -x-\frac{\pi}{5}+2n\pi\,\textrm{.}\end{align}\right.</math>}} |
If we collect ''x'' onto one side, we end up with | If we collect ''x'' onto one side, we end up with | ||
- | {{ | + | {{Abgesetzte Formel||<math>\left\{\begin{align} |
x &= \frac{\pi}{20} + \frac{n\pi}{2}\,,\\[5pt] | x &= \frac{\pi}{20} + \frac{n\pi}{2}\,,\\[5pt] | ||
x &= -\frac{\pi }{30}+\frac{n\pi}{3}\,, | x &= -\frac{\pi }{30}+\frac{n\pi}{3}\,, |
Version vom 09:00, 22. Okt. 2008
For a fixed value of u, an equality of the form
\displaystyle \cos u=\cos v |
is satisfied by two angles v in the unit circle,
\displaystyle v=u\qquad\text{and}\qquad v=-u\,\textrm{.} |
This means that all angles v which satisfy the equality are
\displaystyle v=u+2n\pi\qquad\text{and}\qquad v=-u+2n\pi\,, |
where n is an arbitrary integer.
Therefore, the equation
\displaystyle \cos 5x=\cos (x+\pi/5) |
has the solutions
\displaystyle \left\{\begin{align} 5x&=x+\frac{\pi}{5}+2n\pi\quad\text{or}\\[5pt] 5x &= -x-\frac{\pi}{5}+2n\pi\,\textrm{.}\end{align}\right. |
If we collect x onto one side, we end up with
\displaystyle \left\{\begin{align}
x &= \frac{\pi}{20} + \frac{n\pi}{2}\,,\\[5pt] x &= -\frac{\pi }{30}+\frac{n\pi}{3}\,, \end{align}\right. |
where n is an arbitrary integer.