Solution 4.4:1g - Revision history

Tek at 13:54, 10 October 2008

Tek — Fri, 10 Oct 2008 13:54:29 GMT

←Older revision		Revision as of 13:54, 10 October 2008
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-	A quick sketch of the unit circle and the line	+	A quick sketch of the unit circle and the line <math>y=(-1/\!\sqrt{3})x</math>, which corresponds to the tangent value <math>-1/\!\sqrt{3}</math> shows that there are two angles which satisfy <math>\tan v = -1/\!\sqrt{3}</math>. One of the angles lies in the fourth quadrant and the other is the opposite angle in the second quadrant.
-	<math>y=-\~~frac{1}{~~\sqrt{3}}x</math>, which corresponds to the tangent value	+
-	<math>-{1}/{\sqrt{3}~~}\;~~</math>	+
-	shows that there are two angles which satisfy	+
-	<math>\tan v=-{1}/{\sqrt{3}~~}\;~~</math>. One of the angles lies in the fourth quadrant and the other is the opposite angle in the second quadrant.	+

	[[Image:4_4_1_g1.gif\|center]]		[[Image:4_4_1_g1.gif\|center]]
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	[[Image:4_4_1_g2.gif\|center]]		[[Image:4_4_1_g2.gif\|center]]

-	If we call	+	If we call <math>x</math> the side adjacent to the angle <math>\alpha</math> then the fact that <math>\tan v=-1/\!\sqrt{3}</math> gives the length of the opposite side as <math>x/\!\sqrt{3}</math>.
-	<math>x</math>	+
-	the side adjacent to the angle	+
-	<math>\alpha </math>	+
-	then the fact that	+
-	<math>\tan v=-{1}/{\sqrt{3}~~}\;~~</math>	+
-	gives the length of the opposite side as	+
-	<math>{x}/{\sqrt{3}~~}\;~~</math>.	+
-		+

	[[Image:4_4_1_g3.gif\|center]]		[[Image:4_4_1_g3.gif\|center]]

-	~~Pythagoras'~~ theorem gives	+	The Pythagorean theorem gives
-		+
-		+
-	~~<math>x^{2}+\left( \frac{x}{\sqrt{3}} \right)^{2}=1^{2}</math>~~	+
-		+
-		+
-	~~and this equation has the solution~~	+
-	~~<math>x={\sqrt{3}}/{2}\;</math>, which means that~~	+

		+	{{Displayed math\|\|<math>x^2 + \Bigl(\frac{x}{\sqrt{3}}\Bigr)^2 = 1^2</math>}}

-	<math>~~\cos \alpha~~ =~~\frac{{~~\sqrt{3}}/{2~~}\;}{1}=\frac{\sqrt{3}}{2}~~</math>	+	and this equation has the solution <math>x = \sqrt{3}/2</math>, which means that

-	~~i.e.~~	+	{{Displayed math\|\|<math>\cos\alpha = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2}\,,</math>}}
-	<math>\alpha ={\pi }/{6}\~~;</math>. Because the angle~~	+
-	~~<math>v</math>~~	+
-	~~in the fourth quadrant is the complement of~~	+
-	~~<math>~~\~~alpha~~ </math>	+
-		+
-	~~<math>v</math>~~	+
-	~~is given by~~	+

		+	i.e. <math>\alpha = \pi/6</math>. Because the angle <math>v</math> in the fourth quadrant is the complement of <math>\alpha</math> the angle <math>v</math> is given by

-	<math>v=2\pi -\alpha =2\pi -\frac{\pi }{6}=\frac{11\pi }{6}~~</math>.~~	+	{{Displayed math\|\|<math>v = 2\pi - \alpha = 2\pi - \frac{\pi}{6} = \frac{11\pi }{6}\,\textrm{.}</math>}}
-	~~If we subtract half a turn~~,	+
-	~~<math>~~\pi </math>~~, we obtain the other angle~~	+

		+	If we subtract half a turn, <math>\pi</math>, we obtain the other angle

-	<math>v=\frac{11\pi }{6}-\pi =\frac{5\pi }{6}</math>	+	{{Displayed math\|\|<math>v = \frac{11\pi}{6} - \pi = \frac{5\pi}{6}\,\textrm{.}</math>}}

Ian at 13:00, 30 September 2008

Ian — Tue, 30 Sep 2008 13:00:31 GMT

←Older revision		Revision as of 13:00, 30 September 2008
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-	~~{{NAVCONTENT_START}}~~	+	A quick sketch of the unit circle and the line
-	<~~center~~> ~~[[Image:4_4_1g~~-1~~(2).gif]]~~ </~~center~~>	+	<math>y=-\frac{1}{\sqrt{3}}x</math>, which corresponds to the tangent value
-	{~~{NAVCONTENT_STOP~~}}	+	<math>-{1}/{\sqrt{3}}\;</math>
-	{{~~NAVCONTENT_START~~}}	+	shows that there are two angles which satisfy
-	<~~center~~> ~~[[Image:4_4_1g~~-~~2(2).gif]] <~~/~~center>~~	+	<math>\tan v=-{1}/{\sqrt{3}}\;</math>. One of the angles lies in the fourth quadrant and the other is the opposite angle in the second quadrant.
-	{{~~NAVCONTENT_STOP~~}}	+
-		+

	[[Image:4_4_1_g1.gif\|center]]		[[Image:4_4_1_g1.gif\|center]]
		+
		+	We can therefore limit ourselves to the fourth quadrant and draw an auxiliary triangle in order to determine the angle there.

	[[Image:4_4_1_g2.gif\|center]]		[[Image:4_4_1_g2.gif\|center]]
		+
		+	If we call
		+	<math>x</math>
		+	the side adjacent to the angle
		+	<math>\alpha </math>
		+	then the fact that
		+	<math>\tan v=-{1}/{\sqrt{3}}\;</math>
		+	gives the length of the opposite side as
		+	<math>{x}/{\sqrt{3}}\;</math>.
		+

	[[Image:4_4_1_g3.gif\|center]]		[[Image:4_4_1_g3.gif\|center]]
		+
		+	Pythagoras' theorem gives
		+
		+
		+	<math>x^{2}+\left( \frac{x}{\sqrt{3}} \right)^{2}=1^{2}</math>
		+
		+
		+	and this equation has the solution
		+	<math>x={\sqrt{3}}/{2}\;</math>, which means that
		+
		+
		+	<math>\cos \alpha =\frac{{\sqrt{3}}/{2}\;}{1}=\frac{\sqrt{3}}{2}</math>
		+
		+	i.e.
		+	<math>\alpha ={\pi }/{6}\;</math>. Because the angle
		+	<math>v</math>
		+	in the fourth quadrant is the complement of
		+	<math>\alpha </math>
		+
		+	<math>v</math>
		+	is given by
		+
		+
		+	<math>v=2\pi -\alpha =2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>.
		+	If we subtract half a turn,
		+	<math>\pi </math>, we obtain the other angle
		+
		+
		+	<math>v=\frac{11\pi }{6}-\pi =\frac{5\pi }{6}</math>

Tekbot: Lösning 4.4:1g moved to Solution 4.4:1g: Robot: moved page

Tekbot — Tue, 09 Sep 2008 10:01:43 GMT

Lösning 4.4:1g moved to Solution 4.4:1g: Robot: moved page

←Older revision

Revision as of 10:01, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:17:59 GMT

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

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Madde at 10:51, 19 August 2008

Madde — Tue, 19 Aug 2008 10:51:30 GMT

←Older revision		Revision as of 10:51, 19 August 2008
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Madde at 10:51, 19 August 2008

Madde — Tue, 19 Aug 2008 10:51:00 GMT

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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_4_1g-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_4_1g-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Thu, 03 Apr 2008 12:23:11 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_1g-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_1g-2(2).gif </center> {{NAVCONTENT_STOP}}

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