I need to fine the exact value of these problems. Thank you!
1) cos(2 cos^-1 4/5)
2) cos [2 tan^-1 (-4/3)]
1) cos(2 cos^-1 4/5)
$\displaystyle let\ \cos^{-1} \left (\frac{4}{5}\right )=x \quad \Rightarrow \cos x= \frac{4}{5} \quad ..............(1) $
$\displaystyle \therefore \cos\left (2 \cos^{-1} {\frac{4}{5}}\right ) =\cos (2x)=2\cos^2x-1$
substitute value of cos x from (1) and solve
$\displaystyle let \tan^{-1} \left (\frac{-4}{3}\right )= x \quad \Rightarrow \tan x=\frac{-4}{3} \quad ..............(2) $2) cos [2 tan^-1 (-4/3)]
$\displaystyle \therefore \cos \left \{2 \tan^{-1} \left ( \frac{-4}{3} \right )\right \}= \cos(2x)= \frac{1-tan^2x}{1+tan^2x}$
substitute value of tan x from (2) and solve