Question

$cos[{2cos}^{-1}\frac{1}{5}+{sin}^{-1}\frac{1}{5}]=$

Answer 1

$cos[2co{s}^{-1}\left(\frac{1}{5}\right)+si{n}^{-1}\left(\frac{1}{5}\right)]$

$cos[\frac{\pi }{\alpha }+co{s}^{-1}\left(\frac{1}{5}\right)]$

$-sin\left[co{s}^{-1}\left(\frac{1}{5}\right)\right]\Rightarrow$

$\theta ϵ[0,\pi /2]$

$\theta =co{s}^{-1}\frac{1}{5}$

$cos\theta =\frac{1}{5}$

$sin\theta =\frac{\sqrt[2]{26}}{5}$

so, sin$\left(co{s}^{-1}\frac{1}{5}\right)=\frac{\sqrt[2]{26}}{5}$

Thus = $-sin\left[co{s}^{-1}\left(\frac{1}{5}\right)\right]=\frac{\sqrt[-2]{-26}}{5}$