Lösung 4.3:9
Aus Online Mathematik Brückenkurs 1
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Using the formula for double angles on <math>\sin 160^{\circ}</math> gives | Using the formula for double angles on <math>\sin 160^{\circ}</math> gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}} |
On the right-hand side, we see that the factor <math>\cos 80^{\circ}</math> has appeared, and if we use the formula for double angles on the second factor (<math>\sin 80^{\circ}</math>), | On the right-hand side, we see that the factor <math>\cos 80^{\circ}</math> has appeared, and if we use the formula for double angles on the second factor (<math>\sin 80^{\circ}</math>), | ||
| - | {{ | + | {{Abgesetzte Formel||<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}} |
we obtain a further factor <math>\cos 40^{\circ}</math>. A final application of the formula for double angles on <math>\sin 40^{\circ }</math> gives us all three cosine factors, | we obtain a further factor <math>\cos 40^{\circ}</math>. A final application of the formula for double angles on <math>\sin 40^{\circ }</math> gives us all three cosine factors, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·}</math>}} |
We have thus succeeded in showing that | We have thus succeeded in showing that | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ}</math>}} |
which can also be written as | which can also be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.}</math>}} |
[[Image:4_3_9.gif||right]] | [[Image:4_3_9.gif||right]] | ||
Version vom 08:57, 22. Okt. 2008
Using the formula for double angles on \displaystyle \sin 160^{\circ} gives
| \displaystyle \sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.} |
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}\cdot 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}\cdot 2\cos 40^{\circ}\cdot\sin 40^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\cdot 2\cos 20^{\circ}\sin 20^{\circ}\,\textrm{·} |
We have thus succeeded in showing that
| \displaystyle \sin 160^{\circ} = 8\cos 80^{\circ}\cdot \cos 40^{\circ}\cdot \cos 20^{\circ}\cdot\sin 20^{\circ} |
which can also be written as
| \displaystyle \cos 80^{\circ}\cdot\cos 40^{\circ}\cdot \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}}\,\textrm{.} |
If we draw the unit circle, we see that \displaystyle 160^{\circ} makes an angle of \displaystyle 20^{\circ} with the negative x-axis, and therefore the angles \displaystyle 20^{\circ} and \displaystyle 160^{\circ} have the same y-coordinate in the unit circle, i.e.
This shows that
