Lösung 4.3:4e

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The addition formula for sine gives us that
The addition formula for sine gives us that
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{{Displayed math||<math>\sin\Bigl(v+\frac{\pi}{4}\Bigr) = \sin v\cdot\cos\frac{\pi }{4} + \cos v\cdot\sin\frac{\pi}{4}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin\Bigl(v+\frac{\pi}{4}\Bigr) = \sin v\cdot\cos\frac{\pi }{4} + \cos v\cdot\sin\frac{\pi}{4}\,\textrm{.}</math>}}
Because we know from exercise b that <math>\sin v = \sqrt{1-b^2}</math> we use that
Because we know from exercise b that <math>\sin v = \sqrt{1-b^2}</math> we use that
<math>\cos (\pi/4) = \sin (\pi/4) = 1/\!\sqrt{2}</math> to obtain
<math>\cos (\pi/4) = \sin (\pi/4) = 1/\!\sqrt{2}</math> to obtain
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{{Displayed math||<math>\sin\Bigl(v+\frac{\pi }{4}\Bigr) = \sqrt{1-b^2}\cdot\frac{1}{\sqrt{2}} + b\cdot\frac{1}{\sqrt{2}}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\sin\Bigl(v+\frac{\pi }{4}\Bigr) = \sqrt{1-b^2}\cdot\frac{1}{\sqrt{2}} + b\cdot\frac{1}{\sqrt{2}}\,\textrm{.}</math>}}

Version vom 08:55, 22. Okt. 2008

The addition formula for sine gives us that

\displaystyle \sin\Bigl(v+\frac{\pi}{4}\Bigr) = \sin v\cdot\cos\frac{\pi }{4} + \cos v\cdot\sin\frac{\pi}{4}\,\textrm{.}

Because we know from exercise b that \displaystyle \sin v = \sqrt{1-b^2} we use that \displaystyle \cos (\pi/4) = \sin (\pi/4) = 1/\!\sqrt{2} to obtain

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