Lösung 4.3:4e

Aus Online Mathematik Brückenkurs 1

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The addition formula for sine gives us that
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<center> [[Image:4_3_4e.gif]] </center>
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<math>\sin \left( v+\frac{\pi }{4} \right)=\sin v\centerdot \cos \frac{\pi }{4}+\cos v\centerdot \sin \frac{\pi }{4}.</math>
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Because we know from exercise b that
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<math>\sin v=\sqrt{1-b^{2}}</math>
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we use that
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<math>\cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}</math>
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to obtain
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<math>\sin \left( v+\frac{\pi }{4} \right)=\sqrt{1-b^{2}}\centerdot \frac{1}{\sqrt{2}}+b\centerdot \frac{1}{\sqrt{2}}.</math>

Version vom 11:54, 29. Sep. 2008

The addition formula for sine gives us that


\displaystyle \sin \left( v+\frac{\pi }{4} \right)=\sin v\centerdot \cos \frac{\pi }{4}+\cos v\centerdot \sin \frac{\pi }{4}.


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

\displaystyle \sin \left( v+\frac{\pi }{4} \right)=\sqrt{1-b^{2}}\centerdot \frac{1}{\sqrt{2}}+b\centerdot \frac{1}{\sqrt{2}}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:4e“