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Lösung 4.3:8a

Aus Online Mathematik Brückenkurs 1

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Version vom 15:08, 22. Okt. 2008

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.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:8a“