Solution 4.2:4b - Revision history

Tek at 08:37, 9 October 2008

Tek — Thu, 09 Oct 2008 08:37:27 GMT

←Older revision		Revision as of 08:37, 9 October 2008
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-	We start by subtracting	+	We start by subtracting <math>2\pi</math> from <math>11\pi/3</math>, so that we get an angle between <math>0</math> and <math>2\pi </math>. This doesn't change the cosine value
-	<math>2\pi </math>	+
-	from	+
-	<math>~~\frac{~~11\pi }{3}</math>, so that we get an angle between	+
-	<math>o</math>	+
-	and	+
-	<math>2\pi </math>. This doesn't change the cosine value	+

		+	{{Displayed math\|\|<math>\cos\frac{11\pi}{3} = \cos\Bigl(\frac{11\pi}{3}-2\pi\Bigr) = \cos\frac{5\pi}{3}\,\textrm{.}</math>}}

-	<math>~~\cos \frac{11~~\pi }{3~~}=\cos \left( \frac{11~~\pi ~~}{3}~~-2\pi ~~\right)=\cos \frac{5\pi }{3}~~</math>	+	Then, by rewriting <math>5\pi/3</math> as a sum of <math>\pi</math>- and <math>\pi/2</math>-terms,

		+	{{Displayed math\|\|<math>\frac{5\pi}{3} = \frac{3\pi +\dfrac{3}{2}\pi +\dfrac{1}{2}\pi}{3} = \pi + \frac{\pi}{2} + \frac{\pi}{6}</math>}}

-	~~Then, by rewriting~~	+	we see that <math>5\pi/3</math> is an angle in the fourth quadrant which makes an angle <math>\pi/6</math> with the negative ''y''-axis.
-	<math>~~\frac{~~5\pi }{3}</math>	+
-	~~as a sum of~~	+
-	<math>\pi </~~math>~~	+
-	~~- and~~	+
-	~~<math>\frac{\pi }{2}~~</math>	+
-	-~~terms~~	+

		+	[[Image:4_2_4b1.gif\|\|center]]

-	<math>~~\frac{~~5\pi }{3}=\~~frac{3\pi +\frac{3}{2}\pi +\frac{1}{2}\pi }{3}=\pi +\frac{\pi }{2}+\frac{\pi }{6}~~</math>	+	With the help of an auxiliary triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle
		+	<math>5\pi/3\,</math>.

-	~~we see that~~	+	{\| width="100%"
-	<math>\frac{5\pi }{3}~~</math>~~	+	\|width="50%" align="center"\|[[Image:4_2_4_b2.gif]]
-	~~is an angle in the fourth quadrant which makes an angle~~	+	\|width="50%" align="left"\|<math>\begin{align}\text{opposite} &= 1\cdot\sin\frac{\pi}{6} = \frac{1}{2}\\[5pt] \text{adjacent} &= 1\cdot\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\end{align}</math>
-	~~<math>~~\frac{\pi }{6}</math>	+	\|}
-	~~with the negative~~	+
-	~~<math>y</math>~~	+
-	~~-axis.~~	+

		+	The point has coordinates <math>(1/2,-\sqrt{3}/2)</math> and

-	~~[[Image:4_2_4b1.gif]]~~	+	{{Displayed math\|\|<math>\cos \frac{11\pi}{3} = \cos\frac{5\pi}{3} = \frac{1}{2}\,\textrm{.}</math>}}
-		+
-	~~With the help of a triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle~~	+
-	~~<math>\frac~~{~~5\pi }~~{~~3}</~~math~~> .~~	+
-		+
-		+
-	~~[[Image:4_2_4_b2.gif]]~~	+
-		+
-	~~The point has coordinates~~	+
-	~~<math>\left( \frac{1}{2} \right.,\left. -\frac{\sqrt{3}}{2} \right)</math>~~	+
-	~~and~~	+
-	<math>\cos \frac{11\pi }{3}=\cos \frac{5\pi }{3}=\frac{1}{2}</math>.	+

Ian at 13:05, 28 September 2008

Ian — Sun, 28 Sep 2008 13:05:32 GMT

←Older revision		Revision as of 13:05, 28 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	We start by subtracting
-	<~~center~~> ~~[[Image:4_2_4b-1(~~2~~).gif]]~~ </~~center~~>	+	<math>2\pi </math>
-	{{~~NAVCONTENT_STOP~~}}	+	from
-	{{~~NAVCONTENT_START~~}}	+	<math>\frac{11\pi }{3}</math>, so that we get an angle between
-	<~~center~~> ~~[[Image:4_2_4b~~-2(2~~).gif]]~~ </~~center~~>	+	<math>o</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
		+	<math>2\pi </math>. This doesn't change the cosine value
		+
		+
		+	<math>\cos \frac{11\pi }{3}=\cos \left( \frac{11\pi }{3}-2\pi \right)=\cos \frac{5\pi }{3}</math>
		+
		+
		+	Then, by rewriting
		+	<math>\frac{5\pi }{3}</math>
		+	as a sum of
		+	<math>\pi </math>
		+	- and
		+	<math>\frac{\pi }{2}</math>
		+	-terms
		+
		+
		+	<math>\frac{5\pi }{3}=\frac{3\pi +\frac{3}{2}\pi +\frac{1}{2}\pi }{3}=\pi +\frac{\pi }{2}+\frac{\pi }{6}</math>
		+
		+	we see that
		+	<math>\frac{5\pi }{3}</math>
		+	is an angle in the fourth quadrant which makes an angle
		+	<math>\frac{\pi }{6}</math>
		+	with the negative
		+	<math>y</math>
		+	-axis.
		+
		+
	[[Image:4_2_4b1.gif]]		[[Image:4_2_4b1.gif]]
		+
		+	With the help of a triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle
		+	<math>\frac{5\pi }{3}</math> .
		+
		+
	[[Image:4_2_4_b2.gif]]		[[Image:4_2_4_b2.gif]]
		+
		+	The point has coordinates
		+	<math>\left( \frac{1}{2} \right.,\left. -\frac{\sqrt{3}}{2} \right)</math>
		+	and
		+	<math>\cos \frac{11\pi }{3}=\cos \frac{5\pi }{3}=\frac{1}{2}</math>.

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

Tekbot — Tue, 09 Sep 2008 09:46:03 GMT

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

←Older revision

Revision as of 09:46, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:10:09 GMT

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

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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
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-	[[~~Bild~~:4_2_4b1.gif]]	+	[[Image:4_2_4b1.gif]]
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Madde at 15:09, 15 August 2008

Madde — Fri, 15 Aug 2008 15:09:28 GMT

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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	[[Bild:4_2_4b1.gif]]		[[Bild:4_2_4b1.gif]]
-	[[Bild:~~4_2_4b2~~.gif]]	+	[[Bild:4_2_4_b2.gif]]

Madde at 15:08, 15 August 2008

Madde — Fri, 15 Aug 2008 15:08:13 GMT

←Older revision		Revision as of 15:08, 15 August 2008
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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_2_4b-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_2_4b-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Thu, 03 Apr 2008 07:59:47 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_4b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_2_4b-2(2).gif </center> {{NAVCONTENT_STOP}}

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