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 \text{cos 2}v\text{ } in terms of \displaystyle \text{cos }v,


\displaystyle \begin{align} & \text{cos 2}v=\cos ^{2}v-\sin ^{2}v=\cos ^{2}v-\left( 1-\cos ^{2}v \right) \\ & =2\cos ^{2}v-1=2b^{2}-1 \\ \end{align}