Processing Math: Done
Antwort 3.3:6
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
| (Der Versionsvergleich bezieht 5 dazwischen liegende Versionen mit ein.) | |||
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{| width="100%" cellspacing="10px" | {| width="100%" cellspacing="10px" | ||
| - | | | + | |Lösungen: |
|width="100%"| <math>z= \left\{\eqalign{&\textstyle\sqrt[\scriptstyle 4]{2}\bigl(\cos\frac{\pi}{8}+i\,\sin\frac{\pi}{8}\bigr)\cr &\textstyle\sqrt[\scriptstyle 4]{2}\bigl(\cos\frac{9\pi}{8}+i\,\sin\frac{9\pi}{8}\bigr)}\right. = \left\{\eqalign{&\textstyle\phantom{-}{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}+2}\vphantom{2^{2^{\scriptstyle 2}}}}+i\,{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}-2}\vphantom{2^{2^{\scriptstyle 2}}}}\cr &\textstyle -{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}+2}\vphantom{2^{2^{\scriptstyle 2}}}}-i\,{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}-2}\vphantom{2^{2^{\scriptstyle 2}}}}}\right.</math> | |width="100%"| <math>z= \left\{\eqalign{&\textstyle\sqrt[\scriptstyle 4]{2}\bigl(\cos\frac{\pi}{8}+i\,\sin\frac{\pi}{8}\bigr)\cr &\textstyle\sqrt[\scriptstyle 4]{2}\bigl(\cos\frac{9\pi}{8}+i\,\sin\frac{9\pi}{8}\bigr)}\right. = \left\{\eqalign{&\textstyle\phantom{-}{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}+2}\vphantom{2^{2^{\scriptstyle 2}}}}+i\,{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}-2}\vphantom{2^{2^{\scriptstyle 2}}}}\cr &\textstyle -{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}+2}\vphantom{2^{2^{\scriptstyle 2}}}}-i\,{\textstyle\frac{1}{2}}\sqrt{\smash{2\sqrt{2}-2}\vphantom{2^{2^{\scriptstyle 2}}}}}\right.</math> | ||
|- | |- | ||
| - | | | + | |Ausdruck: |
|width="100%"| <math>\displaystyle\tan \frac{\pi}{8} = \sqrt{2} - 1</math> | |width="100%"| <math>\displaystyle\tan \frac{\pi}{8} = \sqrt{2} - 1</math> | ||
|} | |} | ||
Aktuelle Version
| Lösungen: |     42 cos 8+isin8 42cos89+isin89= 21 22+2+i2122−2−2122+2−i2122−2 |
| Ausdruck: |  8=2−1 |
