Solution 4.3:6a - Revision history

Tek at 07:35, 10 October 2008

2008-10-10T07:35:06Z

←Older revision		Revision as of 07:35, 10 October 2008
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-	If we think of the angle v as an angle in the unit circle, then	+	If we think of the angle v as an angle in the unit circle, then ''v'' lies in the fourth quadrant and has ''x''-coordinate 3/4.
-	~~<math>~~v~~</math>~~	+
-	lies in the fourth quadrant and has	+
-	~~<math>~~x~~</math>~~	+
-	-coordinate	+
-	~~<math>\frac{~~3}{4~~}</math>~~.	+
-		+

	[[Image:4_3_6_a1.gif\|center]]		[[Image:4_3_6_a1.gif\|center]]

-	If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to	+	If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to 1 and an opposite side equal to 3/4.
-	~~<math>\text{~~1~~}</math>~~	+
-	and an opposite side equal to	+
-	~~<math>\frac{~~3}{4~~}</math>~~.	+

	[[Image:4_3_6_a2.gif\|center]]		[[Image:4_3_6_a2.gif\|center]]

-	Using ~~Pythagoras'~~ theorem, it is possible to determine the remaining side from	+	Using the Pythagorean theorem, it is possible to determine the remaining side from
-		+
-		+
-		+
-	~~<math>b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2}</math>~~	+

		+	{{Displayed math\|\|<math>b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2</math>}}

	which gives that		which gives that

		+	{{Displayed math\|\|<math>b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}</math>}}

-	~~<math>\begin{align}~~	+	Because the angle ''v'' belongs to the fourth quadrant, its ''y''-coordinate is negative and is therefore equal to <math>-b</math>, i.e.
-	~~& b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2} \\~~	+
-	~~& b=\sqrt{1-\left( \frac{3}{4} \right)^{2}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	Because the angle	+
-	~~<math>~~v~~</math>~~	+
-	belongs to the fourth quadrant, its	+
-	~~<math>~~y~~</math>~~	+
-	-coordinate is negative and is therefore equal to	+
-	<math>-b</math>, i.e.	+
-		+
-		+
-	~~<math>\sin v=-\frac{\sqrt{7}}{4}</math>~~	+

		+	{{Displayed math\|\|<math>\sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}</math>}}

	Thus, we have directly that		Thus, we have directly that

-		+	{{Displayed math\|\|<math>\tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}</math>}}
-	<math>\tan v=\frac{\sin v}{\cos v}=\frac{-~~\frac{~~\sqrt{7}}{4}}~~{\frac~~{3}{4}}=-\frac{\sqrt{7}}{3}</math>	+

Ian at 09:12, 30 September 2008

2008-09-30T09:12:56Z

←Older revision		Revision as of 09:12, 30 September 2008
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-	~~{{NAVCONTENT_START}}~~	+	If we think of the angle v as an angle in the unit circle, then
-	<~~center~~> ~~[[Image:4_3_6a-1(2).gif]]~~ </~~center~~>	+	<math>v</math>
-	~~{{NAVCONTENT_STOP}}~~	+	lies in the fourth quadrant and has
-	~~{{NAVCONTENT_START}}~~	+	<math>x</math>
-	<~~center~~> ~~[[Image:4_3_6a-2(2).gif]]~~ </~~center~~>	+	-coordinate
-	{{~~NAVCONTENT_STOP}~~}	+	<math>\frac{3}{4}</math>.
		+

	[[Image:4_3_6_a1.gif\|center]]		[[Image:4_3_6_a1.gif\|center]]
		+
		+	If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to
		+	<math>\text{1}</math>
		+	and an opposite side equal to
		+	<math>\frac{3}{4}</math>.
		+
	[[Image:4_3_6_a2.gif\|center]]		[[Image:4_3_6_a2.gif\|center]]
		+
		+	Using Pythagoras' theorem, it is possible to determine the remaining side from
		+
		+
		+
		+	<math>b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2}</math>
		+
		+
		+	which gives that
		+
		+
		+	<math>\begin{align}
		+	& b^{2}+\left( \frac{3}{4} \right)^{2}=1^{2} \\
		+	& b=\sqrt{1-\left( \frac{3}{4} \right)^{2}}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4} \\
		+	\end{align}</math>
		+
		+
		+	Because the angle
		+	<math>v</math>
		+	belongs to the fourth quadrant, its
		+	<math>y</math>
		+	-coordinate is negative and is therefore equal to
		+	<math>-b</math>, i.e.
		+
		+
		+	<math>\sin v=-\frac{\sqrt{7}}{4}</math>
		+
		+
		+	Thus, we have directly that
		+
		+
		+	<math>\tan v=\frac{\sin v}{\cos v}=\frac{-\frac{\sqrt{7}}{4}}{\frac{3}{4}}=-\frac{\sqrt{7}}{3}</math>

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

2008-09-09T09:56:23Z

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

←Older revision

Revision as of 09:56, 9 September 2008

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

2008-08-21T07:15:19Z

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

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-	[[~~Bild~~:4_3_6_a1.gif\|center]]	+	[[Image:4_3_6_a1.gif\|center]]
-	[[~~Bild~~:4_3_6_a2.gif\|center]]	+	[[Image:4_3_6_a2.gif\|center]]

Madde at 14:51, 18 August 2008

2008-08-18T14:51:38Z

←Older revision		Revision as of 14:51, 18 August 2008
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2008-04-03T10:01:50Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_6a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_3_6a-2(2).gif </center> {{NAVCONTENT_STOP}}

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