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Find exact value of cos(2 tan^-1(12/5))

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\bf cos\left[2\ tan^(-1)\left( (12)/(5) \right) \right] \\\\\\ \textit{let's say, hmm }tan^(-1)\left( (12)/(5)\right)=\theta\textit{ that simply means} \\\\\\ cos\left[2\ tan^(-1)\left( (12)/(5) \right) \right]\iff cos(2\theta)\implies 2cos^2(\theta)-1\leftarrow \begin{array}{llll} using\\ double\\angle\\ identities \end{array}


\bf \textit{so hmm what is the }cos(\theta)?\quad cos(\theta)=\cfrac{adjacent}{hypotenuse} \\\\\\ \textit{so, let's use the pythagorean theorem, to get the hypotenuse} \\\\\\ tan^(-1)\left( (12)/(5)\right)=\theta\implies tan(\theta)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a} \\\\\\ c^2=a^2+b^2\implies c=√(a^2+b^2)\implies c=√(5^2+12^2)\implies c=√(169) \\\\\\ c=13


\bf 2cos^2(\theta)-1\iff 2\left[ cos(\theta) \right]^2-1\implies 2\left[ \cfrac{5}{13} \right]^2-1 \\\\\\ 2\cdot \cfrac{25}{169}-1\implies \cfrac{50}{169}-1\implies -\cfrac{119}{169}
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User Ahmed Musallam
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