Solution 4.2:4a - Revision history

Tek at 08:27, 9 October 2008

Tek — Thu, 09 Oct 2008 08:27:04 GMT

←Older revision		Revision as of 08:27, 9 October 2008
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-	It can be a little difficult to draw the angle	+	It can be a little difficult to draw the angle <math>11\pi/6</math> straight onto a unit circle, but if we rewrite <math>11\pi/6</math> as
-	<math>~~\frac{~~11\pi }{6}</math>	+
-	straight onto a unit circle, but if we rewrite	+
-	<math>~~\frac{~~11\pi }{6}</math>	+
-		+
-	as	+
-		+
-	~~<math>\frac{11\pi }{6}=\frac{6\pi +3\pi +2\pi }{6}=\pi +\frac{\pi }{2}+\frac{\pi }{3}</math>~~	+

		+	{{Displayed math\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}

	we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.		we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.

-	We also note that this angle corresponds to exactly the same point on the unit circle as the angle	+	We also note that this angle corresponds to exactly the same point on the unit circle as the angle <math>-\pi/6</math>, and because we calculated <math>\cos (-\pi/6)</math> in exercise 4.2:3f, we have that
-	<math>-~~\frac{~~\pi }{6}</math>, and because we calculated	+
-	<math>\cos ~~\left~~( -~~\frac{~~\pi }{6~~} \right~~)</math>	+
-	in exercise f, we have that	+
-		+
-		+
-	~~<math>\cos \frac{11\pi }{6}=\cos \left( -\frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}</math>~~	+

		+	{{Displayed math\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}

	[[Image:4_2_4_a.gif\|center]]		[[Image:4_2_4_a.gif\|center]]

Ian at 12:52, 28 September 2008

Ian — Sun, 28 Sep 2008 12:52:21 GMT

←Older revision		Revision as of 12:52, 28 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	It can be a little difficult to draw the angle
-	<~~center~~> ~~[[Image:4_2_4a~~.~~gif]]~~ </~~center~~>	+	<math>\frac{11\pi }{6}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	straight onto a unit circle, but if we rewrite
		+	<math>\frac{11\pi }{6}</math>
		+
		+	as
		+
		+	<math>\frac{11\pi }{6}=\frac{6\pi +3\pi +2\pi }{6}=\pi +\frac{\pi }{2}+\frac{\pi }{3}</math>
		+
		+
		+	we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.
		+
		+	We also note that this angle corresponds to exactly the same point on the unit circle as the angle
		+	<math>-\frac{\pi }{6}</math>, and because we calculated
		+	<math>\cos \left( -\frac{\pi }{6} \right)</math>
		+	in exercise f, we have that
		+
		+
		+	<math>\cos \frac{11\pi }{6}=\cos \left( -\frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}</math>
		+
		+
	[[Image:4_2_4_a.gif\|center]]		[[Image:4_2_4_a.gif\|center]]

Tekbot: Lösning 4.2:4a moved to Solution 4.2:4a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:45:43 GMT

Lösning 4.2:4a moved to Solution 4.2:4a: Robot: moved page

←Older revision

Revision as of 09:45, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:09:59 GMT

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

←Older revision		Revision as of 07:09, 21 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_4a.gif]] </center>	+	<center> [[Image:4_2_4a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	[[~~Bild~~:4_2_4_a.gif\|center]]	+	[[Image:4_2_4_a.gif\|center]]

Madde at 15:05, 15 August 2008

Madde — Fri, 15 Aug 2008 15:05:55 GMT

←Older revision		Revision as of 15:05, 15 August 2008
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	<center> [[Bild:4_2_4a.gif]] </center>		<center> [[Bild:4_2_4a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+	[[Bild:4_2_4_a.gif\|center]]

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_2_4a.gif {{NAVCONTENT_STOP}}

Oskar — Thu, 03 Apr 2008 07:58:55 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_4a.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:4_2_4a.gif]] </center>
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