Solution 4.2:3f - Revision history

Tek at 08:21, 9 October 2008

Tek — Thu, 09 Oct 2008 08:21:47 GMT

←Older revision		Revision as of 08:21, 9 October 2008
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-	The point on the unit circle which corresponds to the angle	+	The point on the unit circle which corresponds to the angle <math>-\pi/6</math> lies in the fourth quadrant.
-	<math>-{\pi }/{6~~}\;~~</math>	+
-	lies in the fourth quadrant.	+

-	[[Image:4_2_3_f1.gif]]	+	[[Image:4_2_3_f1.gif\|\|center]]

-	As usual,	+	As usual, <math>\cos (-\pi/6)</math> gives the ''x''-coordinate of the point of intersection between the angle's line and the unit circle. In order to determine this point, we introduce an auxiliary triangle in the fourth quadrant.
-	<math>\cos ~~\left~~( -{\pi }/{6~~}\; \right~~)</math>	+
-	gives the	+
-	~~<math>~~x~~</math>~~	+
-	-coordinate of the point of intersection between the angle's line and the unit circle. In order to determine this point, we introduce an auxiliary triangle in the fourth quadrant.	+

-	[[Image:4_2_3_f2.gif]]	+	[[Image:4_2_3_f2.gif\|\|center]]

	We can determine the edges in this triangle by simple trigonometry and then translate these over to the point's coordinates.		We can determine the edges in this triangle by simple trigonometry and then translate these over to the point's coordinates.

-	[[Image:4_2_3_f3.gif]]	+	{\| width="100%"
		+	\|width="50%" align="center"\|[[Image:4_2_3_f3.gif]]
		+	\|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>
		+	\|}

-	The coordinates of the point of intersection are	+	The coordinates of the point of intersection are <math>(\sqrt{3}/2,-1/2)</math> and in particular <math>\cos (-\pi/6) = \sqrt{3}/2\,</math>.
-	<math>~~\left~~( ~~\frac{~~\sqrt{3}}{2~~} \right.~~,~~\left.~~ -~~\frac{~~1}{2~~} \right~~)</math>	+
-	and in particular	+
-	<math>\cos ~~\left~~( -{\pi }/{6~~}\; \right~~)=~~\frac{~~\sqrt{3}}{2}</math>.	+

Ian at 12:35, 28 September 2008

Ian — Sun, 28 Sep 2008 12:35:36 GMT

←Older revision		Revision as of 12:35, 28 September 2008
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-	~~{{NAVCONTENT_START}}~~	+	The point on the unit circle which corresponds to the angle
-	<~~center~~> ~~[[Image:4_2_3f~~-~~1(2).gif]] </center>~~	+	<math>-{\pi }/{6}\;</math>
-	{{~~NAVCONTENT_STOP~~}}	+	lies in the fourth quadrant.
-	{{~~NAVCONTENT_START~~}}	+
-	~~<center> [[Image:4_2_3f-2(2).gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+

	[[Image:4_2_3_f1.gif]]		[[Image:4_2_3_f1.gif]]
		+
		+	As usual,
		+	<math>\cos \left( -{\pi }/{6}\; \right)</math>
		+	gives the
		+	<math>x</math>
		+	-coordinate of the point of intersection between the angle's line and the unit circle. In order to determine this point, we introduce an auxiliary triangle in the fourth quadrant.

	[[Image:4_2_3_f2.gif]]		[[Image:4_2_3_f2.gif]]
		+
		+	We can determine the edges in this triangle by simple trigonometry and then translate these over to the point's coordinates.

	[[Image:4_2_3_f3.gif]]		[[Image:4_2_3_f3.gif]]
		+
		+	The coordinates of the point of intersection are
		+	<math>\left( \frac{\sqrt{3}}{2} \right.,\left. -\frac{1}{2} \right)</math>
		+	and in particular
		+	<math>\cos \left( -{\pi }/{6}\; \right)=\frac{\sqrt{3}}{2}</math>.

Tekbot: Lösning 4.2:3f moved to Solution 4.2:3f: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:45:23 GMT

Lösning 4.2:3f moved to Solution 4.2:3f: 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 07:09:49 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_3f-1(2).gif]] </center>	+	<center> [[Image:4_2_3f-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_3f-2(2).gif]] </center>	+	<center> [[Image:4_2_3f-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[~~Bild~~:4_2_3_f1.gif]]	+	[[Image:4_2_3_f1.gif]]

-	[[~~Bild~~:4_2_3_f2.gif]]	+	[[Image:4_2_3_f2.gif]]

-	[[~~Bild~~:4_2_3_f3.gif]]	+	[[Image:4_2_3_f3.gif]]

Madde at 14:46, 15 August 2008

Madde — Fri, 15 Aug 2008 14:46:08 GMT

←Older revision		Revision as of 14:46, 15 August 2008
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	[[Bild:4_2_3_f1.gif]]		[[Bild:4_2_3_f1.gif]]
		+
	[[Bild:4_2_3_f2.gif]]		[[Bild:4_2_3_f2.gif]]
		+
	[[Bild:4_2_3_f3.gif]]		[[Bild:4_2_3_f3.gif]]

Madde at 14:45, 15 August 2008

Madde — Fri, 15 Aug 2008 14:45:27 GMT

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	<center> [[Bild:4_2_3f-2(2).gif]] </center>		<center> [[Bild:4_2_3f-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
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		+	[[Bild:4_2_3_f1.gif]]
		+	[[Bild:4_2_3_f2.gif]]
		+	[[Bild:4_2_3_f3.gif]]

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_2_3f-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_2_3f-2(2).gif {{NAVCONTENT_STOP}}

Oskar — Thu, 03 Apr 2008 07:57:36 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_3f-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_2_3f-2(2).gif </center> {{NAVCONTENT_STOP}}

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