Solution 4.3:6c - Revision history

Tek at 07:48, 10 October 2008

2008-10-10T07:48:30Z

←Older revision		Revision as of 07:48, 10 October 2008
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-	Because the angle	+	Because the angle <math>v</math> satisfies <math>\pi \le v\le 3\pi/2\,</math>, <math>v</math> belongs to the third quadrant in the unit circle. Furthermore, <math>\tan v = 3</math> gives that the line which corresponds to the angle
-	<math>v</math>	+	<math>v</math> has slope 3.
-	satisfies	+
-	<math>\pi \le v\le ~~\frac{~~3\pi }{2}</math>,	+
-	<math>v</math>	+
-	belongs to the third quadrant in the unit circle. Furthermore,	+
-	<math>\~~text{~~tan }v=~~\text{~~3 }</math>	+
-	gives that the line which corresponds to the angle	+
-	<math>v</math>	+
-		+
-	~~<math>v</math>~~	+
-	has ~~a gradient of~~	+
-	~~<math>\text{~~3~~}</math>~~.	+
-		+

	[[Image:4_3_6_c1.gif\|center]]		[[Image:4_3_6_c1.gif\|center]]

-	~~slope 3~~	+	In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is 1 and the sides have a 3:1 ratio.
-		+
-		+
-	In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is	+
-	~~<math>\text{~~1~~}</math>~~	+
-	and the sides have a	+
-	~~<math>\text{~~3}:~~\text{~~1 ~~}</math>~~	+
-	ratio.	+

	[[Image:4_3_6_c2.gif\|center]]		[[Image:4_3_6_c2.gif\|center]]

-	If we now use ~~Pythagoras'~~ theorem on the triangle, we see that the horizontal side	+	If we now use the Pythagorean theorem on the triangle, we see that the horizontal side ''a'' satisfies
-	~~<math>\text{~~a~~}</math>~~	+
-	satisfies	+
-		+
-		+
-	~~<math>a^{2}+\left( 3a \right)^{2}=1^{2}</math>~~	+
-		+
-		+
-	~~which gives us that~~	+
-		+
-		+
-	~~<math>10a^{2}=1</math>~~	+
-	~~i.e.~~	+
-	~~<math>a=\frac{1}{\sqrt{10}}</math>~~	+
-		+

-	~~Thus, the angle~~	+	{{Displayed math\|\|<math>a^2 + (3a)^2 = 1^2</math>}}
-	~~<math>v</math>'s~~	+
-	~~<math>x</math>~~	+
-	~~-coordinate is~~	+
-	~~<math>-\frac~~{1}{~~\sqrt{10}}</~~math>	+
-	~~and~~	+
-	<math>y</math>	+
-	~~-coordinate is~~	+
-	~~<math>-\frac{3~~}~~{\sqrt{10~~}~~}</math>, i.e.~~	+

-	<math>~~\cos v=--\frac~~{1}{\sqrt{10}}</math>	+	which gives us that <math>10a^{2}=1</math> i.e. <math>a = 1/\!\sqrt{10}\,\textrm{.}</math>

		+	Thus, the angle ''v'''s ''x''-coordinate is <math>-1/\!\sqrt{10}</math> and ''y''-coordinate is <math>-3/\!\sqrt{10}</math>, i.e.

-	<math>\sin v=-\frac{3}{\sqrt{10}}</math>	+	{{Displayed math\|\|<math>\begin{align}
		+	\cos v &= -\frac{1}{\sqrt{10}}\,,\\[5pt]
		+	\sin v &= -\frac{3}{\sqrt{10}}\,\textrm{.}
		+	\end{align}</math>}}

Ian at 09:38, 30 September 2008

2008-09-30T09:38:59Z

←Older revision		Revision as of 09:38, 30 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	Because the angle
-	<~~center~~> ~~[[Image:4_3_6c-1(2).gif]]~~ </~~center~~>	+	<math>v</math>
-	{~~{NAVCONTENT_STOP~~}}	+	satisfies
-	{~~{NAVCONTENT_START}~~}	+	<math>\pi \le v\le \frac{3\pi }{2}</math>,
-	<~~center~~> ~~[[Image:4_3_6c-2(2).gif]]~~ </~~center~~>	+	<math>v</math>
-	{~~{NAVCONTENT_STOP}~~}	+	belongs to the third quadrant in the unit circle. Furthermore,
		+	<math>\text{tan }v=\text{3 }</math>
		+	gives that the line which corresponds to the angle
		+	<math>v</math>
		+
		+	<math>v</math>
		+	has a gradient of
		+	<math>\text{3}</math>.
		+

	[[Image:4_3_6_c1.gif\|center]]		[[Image:4_3_6_c1.gif\|center]]
		+
		+	slope 3
		+
		+
		+	In the third quadrant, we can introduce a right-angled triangle in which the hypotenuse is
		+	<math>\text{1}</math>
		+	and the sides have a
		+	<math>\text{3}:\text{1 }</math>
		+	ratio.
		+
	[[Image:4_3_6_c2.gif\|center]]		[[Image:4_3_6_c2.gif\|center]]
		+
		+	If we now use Pythagoras' theorem on the triangle, we see that the horizontal side
		+	<math>\text{a}</math>
		+	satisfies
		+
		+
		+	<math>a^{2}+\left( 3a \right)^{2}=1^{2}</math>
		+
		+
		+	which gives us that
		+
		+
		+	<math>10a^{2}=1</math>
		+	i.e.
		+	<math>a=\frac{1}{\sqrt{10}}</math>
		+
		+
		+	Thus, the angle
		+	<math>v</math>'s
		+	<math>x</math>
		+	-coordinate is
		+	<math>-\frac{1}{\sqrt{10}}</math>
		+	and
		+	<math>y</math>
		+	-coordinate is
		+	<math>-\frac{3}{\sqrt{10}}</math>, i.e.
		+
		+	<math>\cos v=--\frac{1}{\sqrt{10}}</math>
		+
		+
		+	<math>\sin v=-\frac{3}{\sqrt{10}}</math>

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

2008-09-09T09:57:03Z

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

←Older revision

Revision as of 09:57, 9 September 2008

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

2008-08-21T07:15:39Z

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

←Older revision		Revision as of 07:15, 21 August 2008
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-	<center> [[~~Bild~~:4_3_6c-1(2).gif]] </center>	+	<center> [[Image:4_3_6c-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
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-	[[~~Bild~~:4_3_6_c1.gif\|center]]	+	[[Image:4_3_6_c1.gif\|center]]
-	[[~~Bild~~:4_3_6_c2.gif\|center]]	+	[[Image:4_3_6_c2.gif\|center]]

Madde at 15:52, 18 August 2008

2008-08-18T15:52:30Z

←Older revision		Revision as of 15:52, 18 August 2008
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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[Bild:4_3_6_c1.gif]]	+	[[Bild:4_3_6_c1.gif\|center]]
-	[[Bild:4_3_6_c2.gif]]	+	[[Bild:4_3_6_c2.gif\|center]]

Madde at 15:51, 18 August 2008

2008-08-18T15:51:38Z

←Older revision		Revision as of 15:51, 18 August 2008
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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
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		+	[[Bild:4_3_6_c1.gif]]
		+	[[Bild:4_3_6_c2.gif]]

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_3_6c-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_3_6c-2(2).gif {{NAVCONTENT_STOP}}

2008-04-03T10:03:40Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_6c-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_3_6c-2(2).gif </center> {{NAVCONTENT_STOP}}

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