Solution 4.3:8a - Revision history

Tek at 08:16, 10 October 2008

2008-10-10T08:16:57Z

←Older revision		Revision as of 08:16, 10 October 2008
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-		+	We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that
-		+
-	We rewrite	+
-	<math>\~~text{~~tan }v</math>	+
-	on the left-hand side as	+
-	<math>\frac{\sin v}{\cos v}</math>, so that	+
-		+
-		+
-	~~<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}</math>~~	+

		+	{{Displayed math\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}

	If we then use the Pythagorean identity		If we then use the Pythagorean identity

		+	{{Displayed math\|\|<math>\cos^2\!v + \sin^2\!v = 1</math>}}

-	~~<math>\cos ^{2}v+\sin ^{2}v=1</math>~~	+	and rewrite <math>\cos^2\!v</math> in the denominator as <math>1 - \sin^2\!v</math>, we get what we are looking for on the right-hand side. The whole calculation is
-		+
-		+
-	and rewrite	+
-	<math>\~~text{~~cos}^~~{\text{~~2}}v</math>	+
-	in the denominator as	+
-	<math>~~\text{~~1}-\~~text{~~sin}^{\~~text{2}}~~v~~\text{ }~~</math>, we get what we are looking for on the right-hand side. The whole calculation is	+
-		+

-	<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}</math>	+	{{Displayed math\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

2008-09-30T10:37:25Z

←Older revision		Revision as of 10:37, 30 September 2008
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-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:4_3_8a.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	We rewrite
		+	<math>\text{tan }v</math>
		+	on the left-hand side as
		+	<math>\frac{\sin v}{\cos v}</math>, so that
		+
		+
		+	<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}</math>
		+
		+
		+	If we then use the Pythagorean identity
		+
		+
		+	<math>\cos ^{2}v+\sin ^{2}v=1</math>
		+
		+
		+	and rewrite
		+	<math>\text{cos}^{\text{2}}v</math>
		+	in the denominator as
		+	<math>\text{1}-\text{sin}^{\text{2}}v\text{ }</math>, we get what we are looking for on the right-hand side. The whole calculation is
		+
		+
		+	<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}</math>

2008-09-09T09:58:03Z

←Older revision

Revision as of 09:58, 9 September 2008

2008-08-21T07:16:09Z

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

2008-04-03T10:54:46Z

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