[tex]\bf cos\left[2\ tan^{-1}\left( \frac{12}{5} \right) \right]
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\textit{let's say, hmm }tan^{-1}\left( \frac{12}{5}\right)=\theta\textit{ that simply means}
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cos\left[2\ tan^{-1}\left( \frac{12}{5} \right) \right]\iff cos(2\theta)\implies 2cos^2(\theta)-1\leftarrow
\begin{array}{llll}
using\\
double\\angle\\
identities
\end{array}[/tex]
[tex]\bf \textit{so hmm what is the }cos(\theta)?\quad cos(\theta)=\cfrac{adjacent}{hypotenuse}
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\textit{so, let's use the pythagorean theorem, to get the hypotenuse}
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tan^{-1}\left( \frac{12}{5}\right)=\theta\implies tan(\theta)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}
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c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\implies c=\sqrt{5^2+12^2}\implies c=\sqrt{169}
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c=13[/tex]
[tex]\bf 2cos^2(\theta)-1\iff 2\left[ cos(\theta) \right]^2-1\implies 2\left[ \cfrac{5}{13} \right]^2-1
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2\cdot \cfrac{25}{169}-1\implies \cfrac{50}{169}-1\implies -\cfrac{119}{169}[/tex]
