Solution 4.4:5b - Revision history

Tek at 14:02, 13 October 2008

2008-10-13T14:02:44Z

←Older revision		Revision as of 14:02, 13 October 2008
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	Let's first investigate when the equality		Let's first investigate when the equality

		+	{{Displayed math\|\|<math>\tan u=\tan v</math>}}

-	~~<math>\tan u=\tan v</math>~~	+	is satisfied. Because <math>\tan u</math> can be interpreted as the slope (gradient) of the line which makes an angle ''u'' with the positive ''x''-axis, we see that for a fixed value of <math>\tan u</math>, there are two angles ''v'' in the unit circle with this slope,
-		+
-		+
-	is satisfied. Because	+
-	<math>u</math>	+
-	can be interpreted as the slope (gradient) of the line which makes an angle	+
-	~~<math>~~u~~</math>~~	+
-	with the positive	+
-	~~<math>~~x~~</math>~~	+
-	-axis, we see that for a fixed value of tan u, there are two angles	+
-	~~<math>~~v~~</math>~~	+
-	in the unit circle with this slope:	+
-		+
-		+
-	~~<math>v=u</math>~~	+
-	~~and~~	+
-	~~<math>v=u+\pi </math>~~	+

		+	{{Displayed math\|\|<math>v=u\qquad\text{and}\qquad v=u+\pi\,\textrm{.}</math>}}

	[[Image:4_4_5_b.gif\|center]]		[[Image:4_4_5_b.gif\|center]]

-	slope	+	The angle ''v'' has the same slope after every half turn, so if we add multiples of
-	<math>=\~~text{ tan }u~~</math>	+	<math>\pi</math> to ''u'', we will obtain all the angles ''v'' which satisfy the equality
-	~~slope~~	+
-	~~<math>=\text{ tan }~~u~~</math>~~	+

		+	{{Displayed math\|\|<math>v=u+n\pi\,,</math>}}

-	~~The angle~~	+	where ''n'' is an arbitrary integer.
-	~~<math>v</math>~~	+
-	~~has the same slope after every half turn, so if we add multiples of~~	+
-	~~<math>\pi \text{ }</math>~~	+
-	to	+
-	~~<math>u</math>, we will obtain all the angles~~	+
-	~~<math>v</math>~~	+
-	~~which satisfy the equality~~	+
-		+
-		+
-	~~<math>v=u+n\pi </math>~~	+
-		+
-		+
-	where	+
-	~~<math>~~n~~</math>~~	+
-	is an arbitrary integer.	+

	If we apply this result to the equation		If we apply this result to the equation

-		+	{{Displayed math\|\|<math>\tan x=\tan 4x</math>}}
-	<math>\tan x=\tan 4x</math>	+
-		+

	we see that the solutions are given by		we see that the solutions are given by

		+	{{Displayed math\|\|<math>4x = x+n\pi\qquad\text{(n is an arbitrary integer),}</math>}}

-	~~<math>4x=x+n\pi </math>~~	+	and solving for ''x'' gives
-	(	+
-	~~<math>n</math>~~	+
-	~~an arbitrary integer),~~	+
-		+
-	and solving for	+
-	~~<math>~~x~~</math>~~	+
-	gives	+
-		+

-	<math>x=\~~frac~~{1}{3}n\pi ~~</math>~~	+	{{Displayed math\|\|<math>x = \tfrac{1}{3}n\pi\qquad\text{(n is an arbitrary integer).}</math>}}
-	(	+
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer).	+

Ian at 11:06, 1 October 2008

2008-10-01T11:06:30Z

←Older revision		Revision as of 11:06, 1 October 2008
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-	~~{{NAVCONTENT_START}}~~	+	Let's first investigate when the equality
-	<~~center~~> ~~[[Image:4_4_5b-1~~(2)~~.gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+	<math>\tan u=\tan v</math>
-	<~~center~~> ~~[[Image~~:~~4_4_5b-2(2).gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+
		+	is satisfied. Because
		+	<math>u</math>
		+	can be interpreted as the slope (gradient) of the line which makes an angle
		+	<math>u</math>
		+	with the positive
		+	<math>x</math>
		+	-axis, we see that for a fixed value of tan u, there are two angles
		+	<math>v</math>
		+	in the unit circle with this slope:
		+
		+
		+	<math>v=u</math>
		+	and
		+	<math>v=u+\pi </math>
		+

	[[Image:4_4_5_b.gif\|center]]		[[Image:4_4_5_b.gif\|center]]
		+
		+	slope
		+	<math>=\text{ tan }u</math>
		+	slope
		+	<math>=\text{ tan }u</math>
		+
		+
		+	The angle
		+	<math>v</math>
		+	has the same slope after every half turn, so if we add multiples of
		+	<math>\pi \text{ }</math>
		+	to
		+	<math>u</math>, we will obtain all the angles
		+	<math>v</math>
		+	which satisfy the equality
		+
		+
		+	<math>v=u+n\pi </math>
		+
		+
		+	where
		+	<math>n</math>
		+	is an arbitrary integer.
		+
		+	If we apply this result to the equation
		+
		+
		+	<math>\tan x=\tan 4x</math>
		+
		+
		+	we see that the solutions are given by
		+
		+
		+	<math>4x=x+n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer),
		+
		+	and solving for
		+	<math>x</math>
		+	gives
		+
		+
		+	<math>x=\frac{1}{3}n\pi </math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).

Tekbot: Lösning 4.4:5b moved to Solution 4.4:5b: Robot: moved page

2008-09-09T10:06:03Z

Lösning 4.4:5b moved to Solution 4.4:5b: Robot: moved page

←Older revision

Revision as of 10:06, 9 September 2008

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

2008-08-21T07:20:09Z

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

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-	[[~~Bild~~:4_4_5_b.gif\|center]]	+	[[Image:4_4_5_b.gif\|center]]

Madde at 13:11, 19 August 2008

2008-08-19T13:11:02Z

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Oskar at 12:38, 3 April 2008

2008-04-03T12:38:51Z

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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_4_5b-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_4_52-2(2).gif {{NAVCONTENT_STOP}}

2008-04-03T12:36:16Z

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