Lösung 4.3:9

Aus Online Mathematik Brückenkurs 1

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Using the formula for double angles on sin
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<center> [[Image:4_3_9-1(2).gif]] </center>
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<math>160^{\circ }</math>
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gives
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<center> [[Image:4_3_9-2(2).gif]] </center>
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<math>\sin 160^{\circ }=2\cos 80^{\circ }\sin 80^{\circ }</math>
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On the right-hand side, we see that the factor
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<math>\cos 80^{\circ }</math>
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has appeared, and if we use the formula for double angles on the second factor (
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<math>\sin 80^{\circ }</math>
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),
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<math>2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\sin 40^{\circ }</math>
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we obtain a further factor
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<math>\cos 40^{\circ }</math>. A final application of the formula for double angles on
 +
<math>\sin 40^{\circ }</math>
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gives us all three cosine factors:
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<math>2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot \sin 40^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot 2\cos 20^{\circ }\sin 20^{\circ }</math>
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We have thus succeeded in showing that
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<math>\sin 160^{\circ }=8\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }\sin 20^{\circ }</math>
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which can also be written as
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<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}</math>
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If we draw the unit circle, we see that
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<math>160^{\circ }</math>
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makes an angle of
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<math>20^{\circ }</math>
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with the negative
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<math>x</math>
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-axis, and therefore the angles
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<math>20^{\circ }</math>
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and
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<math>160^{\circ }</math>
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have the same
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<math>y</math>
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-coordinate in the unit circle, i.e.
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<math>\sin 20^{\circ }=\sin 160^{\circ }</math>.
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[[Image:4_3_9.gif|center]]
[[Image:4_3_9.gif|center]]
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This shows that
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<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}</math>

Version vom 11:31, 30. Sep. 2008

Using the formula for double angles on sin \displaystyle 160^{\circ } gives


\displaystyle \sin 160^{\circ }=2\cos 80^{\circ }\sin 80^{\circ }


On the right-hand side, we see that the factor \displaystyle \cos 80^{\circ } has appeared, and if we use the formula for double angles on the second factor ( \displaystyle \sin 80^{\circ } ),


\displaystyle 2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\sin 40^{\circ }


we obtain a further factor \displaystyle \cos 40^{\circ }. A final application of the formula for double angles on \displaystyle \sin 40^{\circ } gives us all three cosine factors:


\displaystyle 2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot \sin 40^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot 2\cos 20^{\circ }\sin 20^{\circ }


We have thus succeeded in showing that


\displaystyle \sin 160^{\circ }=8\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }\sin 20^{\circ }


which can also be written as


\displaystyle \cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}


If we draw the unit circle, we see that \displaystyle 160^{\circ } makes an angle of \displaystyle 20^{\circ } with the negative \displaystyle x -axis, and therefore the angles \displaystyle 20^{\circ } and \displaystyle 160^{\circ } have the same \displaystyle y -coordinate in the unit circle, i.e.

\displaystyle \sin 20^{\circ }=\sin 160^{\circ }.


This shows that


\displaystyle \cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:9“