Solution 4.3:6b - Revision history

Tek at 13:00, 10 October 2008

2008-10-10T13:00:46Z

←Older revision		Revision as of 13:00, 10 October 2008
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-	We draw an angle	+	We draw an angle <math>v</math> in the unit circle, and the fact that <math>\sin v = 3/10</math> means that its ''y''-coordinate equals <math>3/10</math>.
-	<math>v</math>	+
-	in the unit circle, and the fact that	+
-	<math>\~~text{~~sin }v~~\text{ }~~=~~\frac{~~3}{10}</math>	+
-	means that its	+
-	~~<math>~~y~~</math>~~	+
-	-coordinate equals	+
-	<math>~~\frac{~~3}{10}</math>.	+

	[[Image:4_3_6_b1.gif\|center]]		[[Image:4_3_6_b1.gif\|center]]

-	With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of	+	With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of 1 and a vertical side of length 3/10.
-	~~<math>\text{~~1~~}</math>~~	+
-	and a vertical side of length	+
-	~~<math>\frac{~~3}{10~~}</math>~~.	+
-		+
-		+

	[[Image:4_3_6_b2.gif\|center]]		[[Image:4_3_6_b2.gif\|center]]

-	We can determine the triangle's remaining side by using ~~Pythagoras'~~ theorem,	+	We can determine the triangle's remaining side by using the Pythagorean theorem,
-		+
-		+
-	~~<math>a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}</math>~~	+

		+	{{Displayed math\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}

	which gives that		which gives that

		+	{{Displayed math\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}

-	<math>~~a=\sqrt{1~~-~~\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}~~</math>	+	This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have

-	~~This means that the angle's~~	+	{{Displayed math\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}
-	<math~~>x</math>~~	+
-	~~-coordinate is~~	+
-	~~<math>-a</math>, i.e. we have~~	+
-		+
-		+
-	<math>\cos v=-\frac{\sqrt{91}}{10}</math>	+

	and thus		and thus

-		+	{{Displayed math\|\|<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}
-	<math>\tan v=\frac{\sin v}{\cos v}=\frac{\~~frac~~{3}{10}}{-\~~frac~~{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}</math>	+

Ian at 09:15, 10 October 2008

2008-10-10T09:15:43Z

←Older revision		Revision as of 09:15, 10 October 2008
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-	~~{{NAVCONTENT_START}}~~	+	We draw an angle
-	<~~center~~> ~~[[Image:4_3_6b-1(2).gif]]~~ </~~center~~>	+	<math>v</math>
-	{{~~NAVCONTENT_STOP~~}}	+	in the unit circle, and the fact that
-	{~~{NAVCONTENT_START}~~}	+	<math>\text{sin }v\text{ }=\frac{3}{10}</math>
-	<~~center~~> ~~[[Image:4_3_6b-2(2).gif]]~~ </~~center~~>	+	means that its
-	{{~~NAVCONTENT_STOP}~~}	+	<math>y</math>
		+	-coordinate equals
		+	<math>\frac{3}{10}</math>.

	[[Image:4_3_6_b1.gif\|center]]		[[Image:4_3_6_b1.gif\|center]]
		+
		+	With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of
		+	<math>\text{1}</math>
		+	and a vertical side of length
		+	<math>\frac{3}{10}</math>.
		+
		+

	[[Image:4_3_6_b2.gif\|center]]		[[Image:4_3_6_b2.gif\|center]]
		+
		+	We can determine the triangle's remaining side by using Pythagoras' theorem,
		+
		+
		+	<math>a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}</math>
		+
		+
		+	which gives that
		+
		+
		+	<math>a=\sqrt{1-\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}</math>
		+
		+	This means that the angle's
		+	<math>x</math>
		+	-coordinate is
		+	<math>-a</math>, i.e. we have
		+
		+
		+	<math>\cos v=-\frac{\sqrt{91}}{10}</math>
		+
		+	and thus
		+
		+
		+	<math>\tan v=\frac{\sin v}{\cos v}=\frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}</math>

Tekbot: Lösning 4.3:6b moved to Solution 4.3:6b: Robot: moved page

2008-09-09T09:56:43Z

Lösning 4.3:6b moved to Solution 4.3:6b: Robot: moved page

←Older revision

Revision as of 09:56, 9 September 2008

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

2008-08-21T07:15:29Z

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

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-	[[~~Bild~~:4_3_6_b1.gif\|center]]	+	[[Image:4_3_6_b1.gif\|center]]

-	[[~~Bild~~:4_3_6_b2.gif\|center]]	+	[[Image:4_3_6_b2.gif\|center]]

Madde at 15:42, 18 August 2008

2008-08-18T15:42:26Z

←Older revision		Revision as of 15:42, 18 August 2008
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-	[[Bild:4_3_6_b1.gif]]	+	[[Bild:4_3_6_b1.gif\|center]]
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Madde at 15:41, 18 August 2008

2008-08-18T15:41:28Z

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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_3_6b-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_3_6b-2(2).gif {{NAVCONTENT_STOP}}

2008-04-03T10:02:44Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_6b-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_3_6b-2(2).gif </center> {{NAVCONTENT_STOP}}

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