Solution 4.2:5a - Revision history

Tek at 11:01, 9 October 2008

2008-10-09T11:01:24Z

←Older revision		Revision as of 11:01, 9 October 2008
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-	Because	+	Because <math>135^{\circ} = 90^{\circ} + 45^{\circ}</math>, <math>135^{\circ}</math> is an angle in the second quadrant which makes an angle of <math>45^{\circ}</math> with the positive ''y''-axis.
-	<math>~~\text{~~135}^{\circ ~~}\text{~~ }=~~\text{ 9}0~~^{\circ ~~}\text{~~ }+~~\text{~~45}^{\circ }</math>,	+
-	<math>~~\text{~~135}^{\circ ~~}\text{~~ }</math>	+
-	is an angle in the second quadrant which makes an angle of	+
-	<math>~~\text{~~45}^{\circ }</math>	+
-	with the positive	+
-	~~<math>~~y~~</math>~~	+
-	-axis.	+
-		+

	[[Image:4_2_5_a1.gif\|center]]		[[Image:4_2_5_a1.gif\|center]]

-	We can determine the point on the unit circle which corresponds to	+	We can determine the point on the unit circle which corresponds to <math>135^{\circ}</math> by introducing an auxiliary triangle and calculating its edges using trigonometry.
-	<math>~~\text{~~135}^{\circ ~~}\text{~~ }</math>	+
-	by introducing an auxiliary triangle and calculating its edges using trigonometry.	+
-		+
-		+
-	~~[[Image:4_2_5_a2.gif\|center]]~~	+
-		+
-	~~opposite<math>=1\centerdot \sin \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>~~	+
-		+
-	~~adjacent~~	+
-	~~<math>=1\centerdot \cos \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>~~	+

		+	{\| width="100%"
		+	\|width="50%" align="center"\|[[Image:4_2_5_a2.gif\|center]]
		+	\|width="50%" align="left"\|<math>\begin{align}\text{opposite} &= 1\cdot\sin 45^{\circ} = \dfrac{1}{\sqrt{2}}\\[5pt] \text{adjacent} &= 1\cdot\cos 45^{\circ} = \frac{1}{\sqrt{2}}\end{align}</math>
		+	\|}

-	The coordinates of the point are	+	The coordinates of the point are <math>( -1/\!\sqrt{2}, 1/\!\sqrt{2})</math> and this shows that <math>\cos 135^{\circ} = -1/\!\sqrt{2}\,</math>.
-	<math>~~\left~~( -\~~frac{1}{~~\sqrt{2}~~} \right.~~,\~~left. \frac{1}{~~\sqrt{2}~~} \right~~)</math>	+
-	and this shows that	+
-	<math>\~~text{cos135}~~^{\circ }=-\~~frac{1}{~~\sqrt{2}}</math>.	+

Ian at 07:56, 29 September 2008

2008-09-29T07:56:26Z

←Older revision		Revision as of 07:56, 29 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Because
-	<~~center~~> ~~[[Image:4_2_5a.gif]]~~ </~~center~~>	+	<math>\text{135}^{\circ }\text{ }=\text{ 9}0^{\circ }\text{ }+\text{45}^{\circ }</math>,
-	{{~~NAVCONTENT_STOP}~~}	+	<math>\text{135}^{\circ }\text{ }</math>
		+	is an angle in the second quadrant which makes an angle of
		+	<math>\text{45}^{\circ }</math>
		+	with the positive
		+	<math>y</math>
		+	-axis.
		+
		+
	[[Image:4_2_5_a1.gif\|center]]		[[Image:4_2_5_a1.gif\|center]]
		+
		+	We can determine the point on the unit circle which corresponds to
		+	<math>\text{135}^{\circ }\text{ }</math>
		+	by introducing an auxiliary triangle and calculating its edges using trigonometry.
		+
		+
	[[Image:4_2_5_a2.gif\|center]]		[[Image:4_2_5_a2.gif\|center]]
		+
		+	opposite<math>=1\centerdot \sin \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>
		+
		+	adjacent
		+	<math>=1\centerdot \cos \centerdot 45^{\circ }=\frac{1}{\sqrt{2}}</math>
		+
		+
		+	The coordinates of the point are
		+	<math>\left( -\frac{1}{\sqrt{2}} \right.,\left. \frac{1}{\sqrt{2}} \right)</math>
		+	and this shows that
		+	<math>\text{cos135}^{\circ }=-\frac{1}{\sqrt{2}}</math>.

Tekbot: Lösning 4.2:5a moved to Solution 4.2:5a: Robot: moved page

2008-09-09T09:47:43Z

Lösning 4.2:5a moved to Solution 4.2:5a: Robot: moved page

←Older revision

Revision as of 09:47, 9 September 2008

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

2008-08-21T07:10:59Z

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

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Madde at 15:32, 15 August 2008

2008-08-15T15:32:54Z

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2008-04-03T08:03:35Z

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