Lösung 4.3:3d

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Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.
Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.
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{{Displayed math||<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}
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{{Abgesetzte Formel||<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}
and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,
and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,
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{{Displayed math||<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}

Version vom 08:54, 22. Okt. 2008

The expression for the angle \displaystyle \pi/2 - v differs from \displaystyle \pi/2 by as much as \displaystyle -v differs from \displaystyle 0. This means that \displaystyle \pi/2-v makes the same angle with the positive y-axis as \displaystyle -v makes with the positive x-axis.

Image:4_3_3_d-1.gif   Image:4_3_3_d-2.gif
Angle v Angle π/2 - v

Therefore, the angle \displaystyle \pi/2 - v has a y-coordinate which is equal to the x-coordinate for the angle v, i.e.

\displaystyle \sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v

and from exercise c, we know that \displaystyle \cos v = \sqrt{1-a^2}\,,

\displaystyle \sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:3d“