Solution 4.3:4a

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The Pythagorean identity
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<math>\cos ^{2}v+\sin ^{2}v=1</math>
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gives us directly that
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<math>\sin ^{2}v=1-\cos ^{2}v=1-b^{2}</math>

Revision as of 11:37, 29 September 2008

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}

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