Solution 4.3:3c

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With the help of the Pythagorean identity, we can express
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With the help of the Pythagorean identity, we can express <math>\cos v</math> in terms of <math>\sin v</math>,
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<math>\cos v</math>
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in terms of
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<math>\text{sin }v</math>,
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{{Displayed math||<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \cos v = \pm\sqrt{1-\sin^2 v}\,\textrm{.}</math>}}
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<math>\cos ^{2}v+\sin ^{2}v=1</math>
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In addition, we know that the angle <math>v</math> lies between <math>-\pi/2</math>
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and <math>\pi/2</math>, i.e. either in the first or fourth quadrant, where angles always have a positive ''x''-coordinate (cosine value); thus, we can conclude that
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{{Displayed math||<math>\cos v = \sqrt{1-\sin^2 v} = \sqrt{1-a^2}\,\textrm{.}</math>}}
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In addition, we know that the angle
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<math>v</math>
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lies between
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<math>-{\pi }/{2}\;</math>
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and
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<math>{\pi }/{2}\;</math>, i.e. either in the first or fourth quadrant, where angles always have a positive
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<math>x</math>
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-coordinate (cosine value); thus, we can conclude that
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<math>\cos v=\sqrt{1-\text{sin}^{2}\text{ }v}=\sqrt{1-a^{2}}</math>
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Current revision

With the help of the Pythagorean identity, we can express \displaystyle \cos v in terms of \displaystyle \sin v,

\displaystyle \cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \cos v = \pm\sqrt{1-\sin^2 v}\,\textrm{.}

In addition, we know that the angle \displaystyle v lies between \displaystyle -\pi/2 and \displaystyle \pi/2, i.e. either in the first or fourth quadrant, where angles always have a positive x-coordinate (cosine value); thus, we can conclude that

\displaystyle \cos v = \sqrt{1-\sin^2 v} = \sqrt{1-a^2}\,\textrm{.}
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