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Solution 4.3:4a

Revision as of 14:07, 9 October 2008 by Tek (Talk | contribs)

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The Pythagorean identity \displaystyle \cos^2 v + \sin^2 v = 1 gives us directly that

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