Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 09:58, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.3:8a moved to Solution 4.3:8a: Robot: moved page) ← Previous diff		Revision as of 10:37, 30 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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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>

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Revision as of 10:37, 30 September 2008

We rewrite tan v on the left-hand side as sinvcosv, so that

tan2v=sin2vcos2v

If we then use the Pythagorean identity

cos2v+sin2v=1

and rewrite cos2v in the denominator as 1−sin2v , we get what we are looking for on the right-hand side. The whole calculation is

tan2v=sin2vcos2v=sin2v1−sin2v