Lösung 4.3:4f

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in terms of <math>\cos v</math> and <math>\sin v</math>,
in terms of <math>\cos v</math> and <math>\sin v</math>,
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{{Displayed math||<math>\cos\Bigl(v-\frac{\pi}{3}\Bigr) = \cos v\cdot \cos\frac{\pi }{3} + \sin v\cdot \sin\frac{\pi}{3}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\cos\Bigl(v-\frac{\pi}{3}\Bigr) = \cos v\cdot \cos\frac{\pi }{3} + \sin v\cdot \sin\frac{\pi}{3}\,\textrm{.}</math>}}
Since <math>\cos v = b</math> and <math>\sin v = \sqrt{1-b^2}</math> we obtain
Since <math>\cos v = b</math> and <math>\sin v = \sqrt{1-b^2}</math> we obtain
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{{Displayed math||<math>\cos\Bigl(v-\frac{\pi}{3}\Bigr) = b\cdot\frac{1}{2} + \sqrt{1-b^2}\cdot\frac{\sqrt{3}}{2}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\cos\Bigl(v-\frac{\pi}{3}\Bigr) = b\cdot\frac{1}{2} + \sqrt{1-b^2}\cdot\frac{\sqrt{3}}{2}\,\textrm{.}</math>}}

Version vom 08:55, 22. Okt. 2008

Using the addition formula for cosine, we can express \displaystyle \cos (v-\pi/3) in terms of \displaystyle \cos v and \displaystyle \sin v,

\displaystyle \cos\Bigl(v-\frac{\pi}{3}\Bigr) = \cos v\cdot \cos\frac{\pi }{3} + \sin v\cdot \sin\frac{\pi}{3}\,\textrm{.}

Since \displaystyle \cos v = b and \displaystyle \sin v = \sqrt{1-b^2} we obtain

\displaystyle \cos\Bigl(v-\frac{\pi}{3}\Bigr) = b\cdot\frac{1}{2} + \sqrt{1-b^2}\cdot\frac{\sqrt{3}}{2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:4f“