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Solution 4.3:8a

From Förberedande kurs i matematik 1

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Current revision (08:16, 10 October 2008) (edit) (undo)
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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
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We rewrite
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<math>\text{tan }v</math>
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on the left-hand side as
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<math>\frac{\sin v}{\cos v}</math>, so that
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<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}</math>
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{{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
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{{Displayed math||<math>\cos^2\!v + \sin^2\!v = 1</math>}}
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<math>\cos ^{2}v+\sin ^{2}v=1</math>
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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
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and rewrite
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<math>\text{cos}^{\text{2}}v</math>
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in the denominator as
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<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
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<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}</math>
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{{Displayed math||<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

Current revision

We rewrite tanv 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.
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