Lösung 4.3:4d

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With the formula for double angles and the Pythagorean identity \displaystyle \cos^2\!v + \sin^2\!v = 1, we can express \displaystyle \cos 2v in terms of \displaystyle \cos v,

\displaystyle \begin{align}

\cos 2v &= \cos^2\!v - \sin^2\!v\\[5pt] &= \cos^2\!v - (1-\cos^2\!v)\\[5pt] &= 2\cos^2\!v-1\\[5pt] &= 2b^2-1\,\textrm{.} \end{align}