Solution 4.3:9 - Revision history

Tek at 11:31, 10 October 2008

2008-10-10T11:31:48Z

←Older revision		Revision as of 11:31, 10 October 2008
Line 1:		Line 1:
-	Using the formula for double angles on ~~sin~~	+	Using the formula for double angles on <math>\sin 160^{\circ}</math> gives
-	<math>160^{\circ }</math>	+
-	gives	+

		+	{{Displayed math\|\|<math>\sin 160^{\circ} = 2\cos 80^{\circ}\sin 80^{\circ}\,\textrm{.}</math>}}

-	<math>~~\sin 160^{\circ }=2~~\cos 80^{\circ }\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>),

		+	{{Displayed math\|\|<math>2\cos 80^{\circ}\sin 80^{\circ} = 2\cos 80^{\circ}\cdot 2\cos 40^{\circ}\sin 40^{\circ}\,,</math>}}

-	~~On the right-hand side, we see that the factor~~	+	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,
-	~~<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>~~	+
-	),	+
-		+
-		+
-	~~<math>2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 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:	+
-		+
-		+
-	~~<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>~~	+

		+	{{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		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>}}
-	<math>\sin 160^{\circ }=8\cos 80^{\circ }\~~centerdot~~ \cos 40^{\circ }\~~centerdot~~ \cos 20^{\circ }\sin 20^{\circ }</math>	+
-		+

	which can also be written as		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>}}

-	<math>~~\cos 80~~^{\circ }~~\centerdot \cos 40~~^{\circ }~~\centerdot \cos~~ 20^{\circ }~~=\frac{\sin~~ 160^{\circ ~~}}{8\sin 20^{\circ }~~}</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>
-	~~If we draw the unit circle, we see that~~	+
-	<~~math~~>~~160^{\circ }</math>~~	+
-	~~makes an angle of~~	+
-	<math>20^{\circ }~~</math>~~	+
-	~~with the negative~~	+
-	~~<math>x</math>~~	+
-	~~-axis, and therefore the angles~~	+
-	~~<math>20^{~~\~~circ }</math>~~	+
-	~~and~~	+
-	~~<math>~~160^{\circ }~~</math>~~	+
-	~~have the same~~	+
-	~~<math>y</math>~~	+
-	~~-coordinate in the unit circle~~, ~~i.e.~~	+
-		+
-	~~<math>~~\~~sin 20^~~{~~\circ }=\sin 160^{\circ~~ }</math>.	+
-		+
-		+
-	~~[[Image:4_3_9.gif\|~~center]]	+

	This shows that		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>
-	<math>\cos 80^{\circ }~~\centerdot~~ \cos 40^{\circ }~~\centerdot~~ \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}</math>	+

Ian at 11:31, 30 September 2008

2008-09-30T11:31:02Z

←Older revision		Revision as of 11:31, 30 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Using the formula for double angles on sin
-	<~~center~~> ~~[[Image:4_3_9~~-1(2).~~gif]]~~ </~~center~~>	+	<math>160^{\circ }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	gives
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:4_3_9~~-~~2(2).gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\sin 160^{\circ }=2\cos 80^{\circ }\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>
		+	),
		+
		+
		+	<math>2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 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:
		+
		+
		+	<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>
		+
		+
		+	We have thus succeeded in showing that
		+
		+
		+	<math>\sin 160^{\circ }=8\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }\sin 20^{\circ }</math>
		+
		+
		+	which can also be written as
		+
		+
		+	<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}</math>
		+
		+
		+	If we draw the unit circle, we see that
		+	<math>160^{\circ }</math>
		+	makes an angle of
		+	<math>20^{\circ }</math>
		+	with the negative
		+	<math>x</math>
		+	-axis, and therefore the angles
		+	<math>20^{\circ }</math>
		+	and
		+	<math>160^{\circ }</math>
		+	have the same
		+	<math>y</math>
		+	-coordinate in the unit circle, i.e.
		+
		+	<math>\sin 20^{\circ }=\sin 160^{\circ }</math>.
		+

	[[Image:4_3_9.gif\|center]]		[[Image:4_3_9.gif\|center]]
		+
		+	This shows that
		+
		+
		+	<math>\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}</math>

Tekbot: Lösning 4.3:9 moved to Solution 4.3:9: Robot: moved page

2008-09-09T09:59:23Z

Lösning 4.3:9 moved to Solution 4.3:9: Robot: moved page

←Older revision

Revision as of 09:59, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T07:16:49Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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-	[[~~Bild~~:4_3_9.gif\|center]]	+	[[Image:4_3_9.gif\|center]]

Madde at 10:16, 19 August 2008

2008-08-19T10:16:24Z

←Older revision		Revision as of 10:16, 19 August 2008
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_3_9-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_3_9-2(2).gif {{NAVCONTENT_STOP}}

2008-04-03T10:05:46Z

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