Solution 4.4:2a - Revision history

Tek at 14:13, 10 October 2008

Tek — Fri, 10 Oct 2008 14:13:57 GMT

←Older revision		Revision as of 14:13, 10 October 2008
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-	We draw a unit circle and mark on those angles on the circle which have a	+	We draw a unit circle and mark those angles on the circle which have a ''y''-coordinate of <math>\sqrt{3}/2</math>, in order to see which solutions lie between
-	~~<math>~~y~~</math>~~	+	<math>0</math> and <math>2\pi</math>.
-	-coordinate of	+
-	<math>{\sqrt{3}}/{2~~}\;~~</math>, in order to see which solutions lie between	+
-	<math>0</math>	+
-	and	+
-	<math>2\pi </math>.	+
-		+
-		+

	[[Image:4_4_2_a.gif\|center]]		[[Image:4_4_2_a.gif\|center]]

-	In the first quadrant, we recognize	+	In the first quadrant, we recognize <math>x = \pi/3</math> as the angle which has a sine value of <math>\sqrt{3}/2</math> and then we have the reflectionally symmetric solution <math>x = \pi - \pi/3 = 2\pi/3</math> in the second quadrant.
-	<math>x={\pi }/{3~~}\;~~</math>	+
-	as the angle which has a sine value of	+
-	<math>{\sqrt{3}}/{2~~}\;~~</math>	+
-	and then we have the reflectionally symmetric solution	+
-	<math>x=\pi -~~\frac{~~\pi }{3}=~~\frac{~~2\pi }{3}</math>	+
-	in the second quadrant.	+
-		+
-	~~Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of~~	+
-	~~<math>2\pi </math>~~	+
-		+
-		+
-		+
-	~~<math>x=\frac{\pi }{3}+2n\pi </math>~~	+
-	~~and~~	+
-	~~<math>x=\frac{2\pi }{3}+2n\pi </math>~~	+
-		+

-	~~where~~	+	Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of <math>2\pi</math>
-	<math>n</math>	+
-	~~is an arbitrary integer.~~	+

-	~~NOTE: when we write that the complete solution is given by~~	+	{{Displayed math\|\|<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,,</math>}}

		+	where ''n'' is an arbitrary integer.

-	<math>x=\frac{\pi }{3}+2n\pi </math>
-	and
-	<math>x=\frac{2\pi }{3}+2n\pi </math>,

-	~~this means that for every integer~~	+	Note: When we write that the complete solution is given by
-	~~<math>n</math>,~~ we ~~obtain a solution to~~ the ~~equation:~~	+

		+	{{Displayed math\|\|<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,\textrm{,}</math>}}

-	~~<math>\begin{array}{*{35}l}~~	+	this means that for every integer ''n'', we obtain a solution to the equation:
-	n~~=0 & x=\frac{\pi }{3} & x=\frac{2\pi }{3} \\~~	+
-	~~n=-1 & x=\frac{\pi }{3}+\left( -1 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -1 \right)\centerdot 2\pi \\~~	+
-	~~n=1 & x=\frac{\pi }{3}+1\centerdot 2\pi & x=\frac{2\pi }{3}+1\centerdot 2\pi \\~~	+
-	~~n=-2 & x=\frac{\pi }{3}+\left( -2 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -2 \right)\centerdot 2\pi \\~~	+
-	~~n=2 & x=\frac{\pi }{3}+2\centerdot 2\pi & x=\frac{2\pi }{3}+2\centerdot 2\pi \\~~	+
-	~~\end{array}</math>~~	+

-	~~and so on.~~	+	{{Displayed math\|\|<math>\begin{array}{llll}
		+	&n=0:\quad &x=\frac{\pi}{3}\quad &x=\frac{2\pi }{3}\\[5pt]
		+	&n=-1:\quad &x=\frac{\pi}{3}+(-1)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-1)\cdot 2\pi\\[5pt]
		+	&n=1:\quad &x=\frac{\pi}{3}+1\cdot 2\pi\quad &x=\frac{2\pi}{3}+1\cdot 2\pi\\[5pt]
		+	&n=-2:\quad &x=\frac{\pi}{3}+(-2)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-2)\cdot 2\pi\\[5pt]
		+	&n=2:\quad &x=\frac{\pi}{3}+2\cdot 2\pi\quad &x=\frac{2\pi}{3}+2\cdot 2\pi\\[5pt]
		+	&\phantom{n}\vdots &\phantom{x}\vdots &\phantom{x}\vdots
		+	\end{array}</math>}}

Ian at 13:30, 30 September 2008

Ian — Tue, 30 Sep 2008 13:30:27 GMT

←Older revision		Revision as of 13:30, 30 September 2008
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-	~~{{NAVCONTENT_START}}~~	+	We draw a unit circle and mark on those angles on the circle which have a
-	<~~center~~> ~~[[Image:4_4_2a-1(2).gif]]~~ </~~center~~>	+	<math>y</math>
-	{{~~NAVCONTENT_STOP~~}}	+	-coordinate of
-	{~~{NAVCONTENT_START}~~}	+	<math>{\sqrt{3}}/{2}\;</math>, in order to see which solutions lie between
-	<~~center~~> ~~[[Image:4_4_2a-2(~~2~~).gif]]~~ </~~center~~>	+	<math>0</math>
-	~~{{NAVCONTENT_STOP}}~~	+	and
		+	<math>2\pi </math>.
		+


	[[Image:4_4_2_a.gif\|center]]		[[Image:4_4_2_a.gif\|center]]
		+
		+	In the first quadrant, we recognize
		+	<math>x={\pi }/{3}\;</math>
		+	as the angle which has a sine value of
		+	<math>{\sqrt{3}}/{2}\;</math>
		+	and then we have the reflectionally symmetric solution
		+	<math>x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}</math>
		+	in the second quadrant.
		+
		+	Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of
		+	<math>2\pi </math>
		+
		+
		+
		+	<math>x=\frac{\pi }{3}+2n\pi </math>
		+	and
		+	<math>x=\frac{2\pi }{3}+2n\pi </math>
		+
		+
		+	where
		+	<math>n</math>
		+	is an arbitrary integer.
		+
		+	NOTE: when we write that the complete solution is given by
		+
		+
		+	<math>x=\frac{\pi }{3}+2n\pi </math>
		+	and
		+	<math>x=\frac{2\pi }{3}+2n\pi </math>,
		+
		+	this means that for every integer
		+	<math>n</math>, we obtain a solution to the equation:
		+
		+
		+	<math>\begin{array}{*{35}l}
		+	n=0 & x=\frac{\pi }{3} & x=\frac{2\pi }{3} \\
		+	n=-1 & x=\frac{\pi }{3}+\left( -1 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -1 \right)\centerdot 2\pi \\
		+	n=1 & x=\frac{\pi }{3}+1\centerdot 2\pi & x=\frac{2\pi }{3}+1\centerdot 2\pi \\
		+	n=-2 & x=\frac{\pi }{3}+\left( -2 \right)\centerdot 2\pi & x=\frac{2\pi }{3}+\left( -2 \right)\centerdot 2\pi \\
		+	n=2 & x=\frac{\pi }{3}+2\centerdot 2\pi & x=\frac{2\pi }{3}+2\centerdot 2\pi \\
		+	\end{array}</math>
		+
		+	and so on.

Tekbot: Lösning 4.4:2a moved to Solution 4.4:2a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 10:02:03 GMT

Lösning 4.4:2a moved to Solution 4.4:2a: Robot: moved page

←Older revision

Revision as of 10:02, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:18:09 GMT

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

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-	[[~~Bild~~:4_4_2_a.gif\|center]]	+	[[Image:4_4_2_a.gif\|center]]

Madde at 12:14, 19 August 2008

Madde — Tue, 19 Aug 2008 12:14:36 GMT

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Oskar — Thu, 03 Apr 2008 12:25:07 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_2a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_2a-2(2).gif </center> {{NAVCONTENT_STOP}}

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