Question

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

Answer 1

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

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

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