Solution 4.4:4 - Revision history

Tek at 13:42, 13 October 2008

Tek — Mon, 13 Oct 2008 13:42:02 GMT

←Older revision		Revision as of 13:42, 13 October 2008
Line 1:		Line 1:
-	The idea is first to find the general solution to the equation and then to see which angles lie between	+	The idea is first to find the general solution to the equation and then to see which angles lie between <math>0^{\circ}</math> and <math>360^{\circ}\,</math>.
-	<math>0^{\circ }</math>	+
-	and	+
-	<math>360^{\circ }</math>.	+

-	If we start by considering the expression	+	If we start by considering the expression <math>2v+10^{\circ}</math> as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
-	<math>~~\text{2}v~~+~~\text{1}0~~^{\circ ~~}\text{~~ }</math>	+
-	as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is	+

		+	{{Displayed math\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}

-	~~<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }</math>~~	+	There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>
-		+	with the negative ''y''-axis and consequently
-		+
-	There is then a further solution which satisfies	+
-	<math>0^{\circ }\le ~~\text{2}v~~+~~\text{1}0~~^{\circ }\le ~~\text{36}0~~^{\circ }</math>, where	+
-	<math>~~\text{2}v~~+~~\text{1}0~~^{\circ ~~}\text{~~ }</math>	+
-	lies in the third quadrant and makes the same angle with the negative y-axis as	+
-	<math>~~\text{1}00~~^{\circ }</math>	+
-	makes with the positive	+
-	~~<math>~~y~~</math>~~	+
-	-axis, i.e.	+
-	<math>~~\text{2}v~~+~~\text{1}0~~^{\circ ~~}\text{~~ }</math>	+
-	makes an angle	+
-	<math>~~\text{11}0~~^{\circ }-~~\text{9}0~~^{\circ }=~~\text{2}0~~^{\circ ~~}~\text{~~ }</math>	+
-	with the negative	+
-	~~<math>~~y~~</math>~~	+
-	-axis and consequently	+
-		+
-		+
-	~~<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>~~	+
-		+

		+	{{Displayed math\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}

	[[Image:4_4_4.gif\|center]]		[[Image:4_4_4.gif\|center]]
-
-
-	There is then a further solution which satisfies
-	<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
-	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
-	lies in the third quadrant and makes the same angle with the negative y-axis as
-	<math>\text{1}00^{\circ }</math>
-	makes with the positive
-	<math>y</math>
-	-axis, i.e.
-	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
-	makes an angle
-	<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
-	with the negative
-	<math>y</math>
-	-axis and consequently
-
-
-	<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
-
-
-	FIGURE1 FIGURE2

	Now it is easy to write down the general solution,		Now it is easy to write down the general solution,

		+	{{Displayed math\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{and}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}

-	~~<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }+n\centerdot 360^{\circ }</math>~~	+	and if we make ''v'' the subject, we get
-	~~and~~	+
-		+
-	~~<math>\text{2}v+\text{1}0^{\circ }=250^{\circ }+n\centerdot 360^{\circ }</math>~~	+
-		+
-		+
-	and if we make	+
-	~~<math>~~v~~</math>~~	+
-	the subject, we get	+
-		+
-		+
-	~~<math>v=50^{\circ }+n\centerdot 180^{\circ }</math>~~	+
-	~~and~~	+
-		+
-	~~<math>v=120^{\circ }+n\centerdot 180^{\circ }</math>~~	+
-	~~EQ6~~	+
-		+
-	~~For different values of the integers~~	+
-	~~<math>n</math>, we see that the corresponding solutions are:~~	+

		+	{{Displayed math\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{and}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}

-	~~<math>\begin{array}{*{35}l}~~	+	For different values of the integers ''n'', we see that the corresponding solutions are:
-	~~\cdots \cdots & \cdots \cdots & \cdots \cdots \\~~	+
-	n~~=-2 & v=50^{\circ }-2\centerdot 180^{\circ }=-310^{\circ } & v=120^{\circ }-2\centerdot 180^{\circ }=-240^{\circ } \\~~	+
-	~~n=-1 & v=50^{\circ }-1\centerdot 180^{\circ }=-130^{\circ } & v=120^{\circ }-1\centerdot 180^{\circ }=-60^{\circ } \\~~	+
-	~~n=0 & v=50^{\circ }+0\centerdot 180^{\circ }=50^{\circ } & v=120^{\circ }+0\centerdot 180^{\circ }=120^{\circ } \\~~	+
-	~~n=1 & v=50^{\circ }+1\centerdot 180^{\circ }=230^{\circ } & v=120^{\circ }+1\centerdot 180^{\circ }=300^{\circ } \\~~	+
-	~~n=2 & v=50^{\circ }+2\centerdot 180^{\circ }=410^{\circ } & v=120^{\circ }+2\centerdot 180^{\circ }=480^{\circ } \\~~	+
-	~~n=3 & v=50^{\circ }+3\centerdot 180^{\circ }=590^{\circ } & v=120^{\circ }+3\centerdot 180^{\circ }=660^{\circ } \\~~	+
-	~~\cdots \cdots & \cdots \cdots & \cdots \cdots \\~~	+
-	~~\end{array}</math>~~	+


		+	{\| align="center"
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|-
		+	\|align="left"\|<math>n=-2:</math>
		+	\|width="20px"\|
		+	\|align="left"\|<math>v = 50^{\circ} - 2\cdot 180^{\circ} = -310^{\circ}</math>
		+	\|width="20px"\|
		+	\|align="left"\|<math>v = 120^{\circ } - 2\cdot 180^{\circ} = -240^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=-1:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} - 1\cdot 180^{\circ} = -130^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} - 1\cdot 180^{\circ} = -60^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=0:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 0\cdot 180^{\circ} = 50^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 0\cdot 180^{\circ} = 120^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=1:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 1\cdot 180^{\circ} = 230^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 1\cdot 180^{\circ} = 300^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=2:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 2\cdot 180^{\circ} = 410^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 2\cdot 180^{\circ} = 480^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=3:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 3\cdot 180^{\circ} = 590^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 3\cdot 180^{\circ} = 660^{\circ}</math>
		+	\|-
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|}

-	From the table, we see that the solutions that are between
-	<math>0^{\circ }</math>
-	and
-	<math>360^{\circ }</math>
-	are

		+	From the table, we see that the solutions that are between <math>0^{\circ}</math> and <math>360^{\circ}</math> are

-	<math>v=50,\quad v=120^{\circ },\quad v=230^{\circ }~~</math>~~	+	{{Displayed math\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}</math>}}
-	and	+
-	~~<math>~~v=300^{\circ }</math>	+

Ian at 10:20, 1 October 2008

Ian — Wed, 01 Oct 2008 10:20:04 GMT

←Older revision		Revision as of 10:20, 1 October 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START}~~}	+	The idea is first to find the general solution to the equation and then to see which angles lie between
-	<~~center~~> ~~[[Image:4_4_4-~~1~~(3)~~.~~gif]]~~ </~~center~~>	+	<math>0^{\circ }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
-	{{~~NAVCONTENT_START~~}}	+	<math>360^{\circ }</math>.
-	<~~center~~> ~~[[Image:4_4_4~~-~~2(3)~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	If we start by considering the expression
-	{{~~NAVCONTENT_START~~}}	+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
-	<~~center~~> ~~[[Image:4_4_4-3(3).gif]]~~ </~~center~~>	+	as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }</math>
		+
		+
		+	There is then a further solution which satisfies
		+	<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	lies in the third quadrant and makes the same angle with the negative y-axis as
		+	<math>\text{1}00^{\circ }</math>
		+	makes with the positive
		+	<math>y</math>
		+	-axis, i.e.
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	makes an angle
		+	<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
		+	with the negative
		+	<math>y</math>
		+	-axis and consequently
		+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
		+


	[[Image:4_4_4.gif\|center]]		[[Image:4_4_4.gif\|center]]
		+
		+
		+	There is then a further solution which satisfies
		+	<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	lies in the third quadrant and makes the same angle with the negative y-axis as
		+	<math>\text{1}00^{\circ }</math>
		+	makes with the positive
		+	<math>y</math>
		+	-axis, i.e.
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	makes an angle
		+	<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
		+	with the negative
		+	<math>y</math>
		+	-axis and consequently
		+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
		+
		+
		+	FIGURE1 FIGURE2
		+
		+	Now it is easy to write down the general solution,
		+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }+n\centerdot 360^{\circ }</math>
		+	and
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=250^{\circ }+n\centerdot 360^{\circ }</math>
		+
		+
		+	and if we make
		+	<math>v</math>
		+	the subject, we get
		+
		+
		+	<math>v=50^{\circ }+n\centerdot 180^{\circ }</math>
		+	and
		+
		+	<math>v=120^{\circ }+n\centerdot 180^{\circ }</math>
		+	EQ6
		+
		+	For different values of the integers
		+	<math>n</math>, we see that the corresponding solutions are:
		+
		+
		+	<math>\begin{array}{*{35}l}
		+	\cdots \cdots & \cdots \cdots & \cdots \cdots \\
		+	n=-2 & v=50^{\circ }-2\centerdot 180^{\circ }=-310^{\circ } & v=120^{\circ }-2\centerdot 180^{\circ }=-240^{\circ } \\
		+	n=-1 & v=50^{\circ }-1\centerdot 180^{\circ }=-130^{\circ } & v=120^{\circ }-1\centerdot 180^{\circ }=-60^{\circ } \\
		+	n=0 & v=50^{\circ }+0\centerdot 180^{\circ }=50^{\circ } & v=120^{\circ }+0\centerdot 180^{\circ }=120^{\circ } \\
		+	n=1 & v=50^{\circ }+1\centerdot 180^{\circ }=230^{\circ } & v=120^{\circ }+1\centerdot 180^{\circ }=300^{\circ } \\
		+	n=2 & v=50^{\circ }+2\centerdot 180^{\circ }=410^{\circ } & v=120^{\circ }+2\centerdot 180^{\circ }=480^{\circ } \\
		+	n=3 & v=50^{\circ }+3\centerdot 180^{\circ }=590^{\circ } & v=120^{\circ }+3\centerdot 180^{\circ }=660^{\circ } \\
		+	\cdots \cdots & \cdots \cdots & \cdots \cdots \\
		+	\end{array}</math>
		+
		+
		+
		+	From the table, we see that the solutions that are between
		+	<math>0^{\circ }</math>
		+	and
		+	<math>360^{\circ }</math>
		+	are
		+
		+
		+	<math>v=50,\quad v=120^{\circ },\quad v=230^{\circ }</math>
		+	and
		+	<math>v=300^{\circ }</math>

Tekbot: Lösning 4.4:4 moved to Solution 4.4:4: Robot: moved page

Tekbot — Tue, 09 Sep 2008 10:05:23 GMT

Lösning 4.4:4 moved to Solution 4.4:4: Robot: moved page

←Older revision

Revision as of 10:05, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:19:49 GMT

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

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Madde at 13:09, 19 August 2008

Madde — Tue, 19 Aug 2008 13:09:11 GMT

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Oskar — Thu, 03 Apr 2008 12:33:34 GMT

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