Lösung 4.3:4b

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If we once again use the Pythagorean identity we get
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<center> [[Image:4_3_4b.gif]] </center>
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<math>\cos ^{2}v+\sin ^{2}v=1\quad \Leftrightarrow \quad \sin v=\pm \sqrt{1-\cos ^{2}v}</math>
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Because the angle v lies between
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<math>0</math>
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and
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<math>\pi </math>,
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<math>\text{sin }v</math>
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is positive (an angle in the first and second quadrants has a positive
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<math>y</math>
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-coordinate) and therefore
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<math>\sin v=+\sqrt{1-\cos ^{2}v}=\sqrt{1-b^{2}}</math>

Version vom 11:42, 29. Sep. 2008

If we once again use the Pythagorean identity we get


\displaystyle \cos ^{2}v+\sin ^{2}v=1\quad \Leftrightarrow \quad \sin v=\pm \sqrt{1-\cos ^{2}v}


Because the angle v lies between \displaystyle 0 and \displaystyle \pi , \displaystyle \text{sin }v is positive (an angle in the first and second quadrants has a positive \displaystyle y -coordinate) and therefore


\displaystyle \sin v=+\sqrt{1-\cos ^{2}v}=\sqrt{1-b^{2}}

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