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Solution 4.3:9

From Förberedande kurs i matematik 1

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m (Lösning 4.3:9 moved to Solution 4.3:9: Robot: moved page)
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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 (
 +
<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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 +
 
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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
 +
<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.
 +
 
 +
<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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 +
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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>

Revision as of 11:31, 30 September 2008

Using the formula for double angles on sin 160 gives


sin160=2cos80sin80


On the right-hand side, we see that the factor cos80 has appeared, and if we use the formula for double angles on the second factor ( sin80 ),


2cos80sin80=2cos802cos40sin40


we obtain a further factor cos40. A final application of the formula for double angles on sin40 gives us all three cosine factors:


2cos802cos40sin40=2cos802cos402cos20sin20


We have thus succeeded in showing that


sin160=8cos80cos40cos20sin20


which can also be written as


cos80cos40cos20=sin1608sin20


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

sin20=sin160.


This shows that


cos80cos40cos20=sin1608sin20=81

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