Lösung 4.3:4f

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Using the addition formula for cosine, we can express
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<center> [[Image:4_3_4f.gif]] </center>
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<math>\cos \left( v-{\pi }/{3}\; \right)</math>
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in terms of
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<math>\text{cos }v</math>
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and
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<math>\text{sin }v</math>,
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<math>\cos \left( v-\frac{\pi }{3} \right)=\cos v\centerdot \cos \frac{\pi }{3}+\sin v\centerdot \sin \frac{\pi }{3}</math>
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Since
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<math>\text{cos }v=b\text{ }</math>
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and
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<math>\sin v=\sqrt{1-b^{2}}</math>
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we obtain
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<math>\cos \left( v-\frac{\pi }{3} \right)=b\centerdot \frac{1}{2}+\sqrt{1-b^{2}}\centerdot \frac{\sqrt{3}}{2}</math>

Version vom 12:00, 29. Sep. 2008

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


\displaystyle \cos \left( v-\frac{\pi }{3} \right)=\cos v\centerdot \cos \frac{\pi }{3}+\sin v\centerdot \sin \frac{\pi }{3}


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


\displaystyle \cos \left( v-\frac{\pi }{3} \right)=b\centerdot \frac{1}{2}+\sqrt{1-b^{2}}\centerdot \frac{\sqrt{3}}{2}

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