Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 10:05, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_9-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_3_9-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (11:31, 10 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Using the formula for double angles on <math>\sin 160^{\circ}</math> gives
-	<center> [[Bild:4_3_9-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:4_3_9-2(2).gif]] </center>	+	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>),
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<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,
		+
		+	{{Displayed math||<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
		+
		+	{{Displayed math||<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
		+
		+	{{Displayed math||<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]]
		+	If we draw the unit circle, we see that <math>160^{\circ}</math> makes an angle of
		+	<math>20^{\circ}</math> with the negative ''x''-axis, and therefore the angles
		+	<math>20^{\circ}</math> and <math>160^{\circ}</math> have the same ''y''-coordinate in the unit circle, i.e.
		+
		+	<center><math>\sin 20^{\circ} = \sin 160^{\circ}\,\textrm{.}</math></center>
		+
		+	This shows that
		+
		+	<center><math>\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} = \frac{\sin 160^{\circ}}{8\sin 20^{\circ}} = \frac{1}{8}\,\textrm{.}</math></center>

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Current revision

Using the formula for double angles on sin160 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.