Antwort 4.3:6
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
(Translated into English) |
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(Der Versionsvergleich bezieht 2 dazwischen liegende Versionen mit ein.) | |||
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{| width="100%" cellspacing="10px" | {| width="100%" cellspacing="10px" | ||
|a) | |a) | ||
- | |width="100%" | <math>\sin{v}=-\displaystyle \frac{\sqrt{7}}{4}\quad</math> | + | |width="100%" | <math>\sin{v}=-\displaystyle \frac{\sqrt{7}}{4}\quad</math> und <math>\quad\tan{v}=-\displaystyle \frac{\sqrt{7}}{3}\,</math> |
|- | |- | ||
|b) | |b) | ||
- | |width="100%" | <math>\cos{v}=-\displaystyle \frac{\sqrt{91}}{10}\quad</math> | + | |width="100%" | <math>\cos{v}=-\displaystyle \frac{\sqrt{91}}{10}\quad</math> und <math>\quad\tan{v}=-\displaystyle \frac{3}{\sqrt{91}}\,</math> |
|- | |- | ||
|c) | |c) | ||
- | |width="100%" | <math>\sin{v}=-\displaystyle \frac{3}{\sqrt{10}}\quad</math> | + | |width="100%" | <math>\sin{v}=-\displaystyle \frac{3}{\sqrt{10}}\quad</math> und <math>\quad\cos{v}=-\displaystyle \frac{1}{\sqrt{10}}\,</math> |
|} | |} |
Aktuelle Version
a) | \displaystyle \sin{v}=-\displaystyle \frac{\sqrt{7}}{4}\quad und \displaystyle \quad\tan{v}=-\displaystyle \frac{\sqrt{7}}{3}\, |
b) | \displaystyle \cos{v}=-\displaystyle \frac{\sqrt{91}}{10}\quad und \displaystyle \quad\tan{v}=-\displaystyle \frac{3}{\sqrt{91}}\, |
c) | \displaystyle \sin{v}=-\displaystyle \frac{3}{\sqrt{10}}\quad und \displaystyle \quad\cos{v}=-\displaystyle \frac{1}{\sqrt{10}}\, |