Solution 4.3:8b - Revision history

Tek at 08:27, 10 October 2008

Tek — Fri, 10 Oct 2008 08:27:22 GMT

←Older revision		Revision as of 08:27, 10 October 2008
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-		+	Because <math>\tan v = \frac{\sin v}{\cos v}</math>, the left-hand side can be written using <math>\cos v</math> as the common denominator,

-	~~Because~~	+	{{Displayed math\|\|<math>\frac{1}{\cos v} - \tan v = \frac{1}{\cos v} - \frac{\sin v}{\cos v} = \frac{1-\sin v}{\cos v}\,\textrm{.}</math>}}
-	<math>\tan v=\frac{\sin v}{\cos v}~~</math>, the left~~-~~hand side can be written using~~	+
-	~~<math>~~\cos v</math>	+
-	~~as the common denominator:~~	+

		+	Now, we observe that if we multiply top and bottom with <math>1+\sin v</math>, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give <math>1-\sin^2\!v = \cos ^2\!v\,</math>, using the difference of two squares,

-	<math>\frac{1}{\cos v}~~-\tan v~~=\frac{1}{\cos v}-\frac{\sin v}{\~~cos~~ v}=\frac~~{\text~~{1-}\sin v}{\cos v}</math>	+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{1-\sin v}{\cos v}
		+	&= \frac{1-\sin v}{\cos v}\cdot\frac{1+\sin v}{1+\sin v}\\[5pt]
		+	&= \frac{1-\sin^2\!v}{\cos v\,(1+\sin v)}\\[5pt]
		+	&= \frac{\cos^2\!v}{\cos v\,(1+\sin v)}\,\textrm{.}
		+	\end{align}</math>}}

		+	Eliminating <math>\cos v</math> then gives the answer,

-	~~Now, we observe that if we multiply top and bottom by with~~	+	{{Displayed math\|\|<math>\frac{\cos^2\!v}{\cos v\,(1+\sin v)} = \frac{\cos v}{1+\sin v}\,\textrm{.}</math>}}
-	~~<math>\text~~{~~1}+\sin v</math>, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give~~	+
-	~~<math>\text~~{~~1}-\sin ^{2}v\text{ }=\cos ^{2}v</~~math~~>, using the conjugate rule:~~	+
-		+
-		+
-	<math>~~\begin{align}~~	+
-	& \frac~~{\text{1-}\sin v}~~{\cos ~~v}=\frac{\text{1-}\sin v}{\cos v}\centerdot \frac{1+\sin v}{1+\sin v}=\frac{1-\sin~~ ^{2}v}{\cos v\~~left~~( 1+\sin v ~~\right~~)} \\	+
-	& =\frac{\cos ~~^{2}~~v}{~~\cos v\left(~~ 1+\sin v ~~\right)~~}. \\	+
-	~~\end~~{~~align~~}</math>	+
-		+
-		+
-	~~Eliminating~~	+
-	~~<math>\cos v</math>~~	+
-	~~then gives the answer:~~	+
-		+
-		+
-	~~<math>\frac{\cos ^{2~~}v}~~{\cos v\left( 1+\sin v \right)}=\frac{\cos v}{1+\sin v}</math>~~	+

Ian at 10:53, 30 September 2008

Ian — Tue, 30 Sep 2008 10:53:01 GMT

←Older revision		Revision as of 10:53, 30 September 2008
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-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image~~:~~4_3_8b~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	Because
		+	<math>\tan v=\frac{\sin v}{\cos v}</math>, the left-hand side can be written using
		+	<math>\cos v</math>
		+	as the common denominator:
		+
		+
		+	<math>\frac{1}{\cos v}-\tan v=\frac{1}{\cos v}-\frac{\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}</math>
		+
		+
		+	Now, we observe that if we multiply top and bottom by with
		+	<math>\text{1}+\sin v</math>, the denominator will contain the denominator of the right-hand side as a factor and, in addition, the numerator can be simplified to give
		+	<math>\text{1}-\sin ^{2}v\text{ }=\cos ^{2}v</math>, using the conjugate rule:
		+
		+
		+	<math>\begin{align}
		+	& \frac{\text{1-}\sin v}{\cos v}=\frac{\text{1-}\sin v}{\cos v}\centerdot \frac{1+\sin v}{1+\sin v}=\frac{1-\sin ^{2}v}{\cos v\left( 1+\sin v \right)} \\
		+	& =\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}. \\
		+	\end{align}</math>
		+
		+
		+	Eliminating
		+	<math>\cos v</math>
		+	then gives the answer:
		+
		+
		+	<math>\frac{\cos ^{2}v}{\cos v\left( 1+\sin v \right)}=\frac{\cos v}{1+\sin v}</math>

Tekbot: Lösning 4.3:8b moved to Solution 4.3:8b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:58:23 GMT

Lösning 4.3:8b moved to Solution 4.3:8b: Robot: moved page

←Older revision

Revision as of 09:58, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:16:19 GMT

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

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Oskar — Thu, 03 Apr 2008 10:55:53 GMT

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