Solution 4.3:4d

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With the formula for double angles and the Pythagorean identity
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With the formula for double angles and the Pythagorean identity <math>\cos^2\!v + \sin^2\!v = 1</math>, we can express <math>\cos 2v</math> in terms of <math>\cos v</math>,
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<math>\cos ^{2}v+\sin ^{2}v=1</math>, we can express
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<math>\text{cos 2}v\text{ }</math>
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in terms of
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<math>\text{cos }v</math>,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\cos 2v &= \cos^2\!v - \sin^2\!v\\[5pt]
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& \text{cos 2}v=\cos ^{2}v-\sin ^{2}v=\cos ^{2}v-\left( 1-\cos ^{2}v \right) \\
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&= \cos^2\!v - (1-\cos^2\!v)\\[5pt]
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& =2\cos ^{2}v-1=2b^{2}-1 \\
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&= 2\cos^2\!v-1\\[5pt]
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\end{align}</math>
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&= 2b^2-1\,\textrm{.}
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\end{align}</math>}}

Current revision

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}

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