Question

$cos[2co{s}^{-1}\frac{1}{5}+si{n}^{-1}\frac{1}{5}]$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{-2\sqrt{6}}{5}$
• C. $-\frac{1}{5}$
• D. $\frac{\sqrt{6}}{5}$

Answer 1

Answer: (B)

$\frac{-2\sqrt{6}}{5}$

$cos\left[2co{s}^{-1}\right(\frac{1}{5})+si{n}^{-1}(\frac{1}{5}\left)\right]$
$=cos\left[co{s}^{-1}\right(\frac{1}{5})+(co{s}^{-1}\left(\frac{1}{5}\right)+si{n}^{-1}\left(\frac{1}{5}\right)\left)\right]$
$=cos\left[co{s}^{-1}\right(\frac{1}{5})+\frac{\pi }{2}]$
$=-sin\left(co{s}^{-1}\right(\frac{1}{5}\left)\right)$
$=-sin\left(si{n}^{-1}\right(\sqrt{1-(\frac{1}{5}{)}^2}\left)\right)$
$=-\sqrt{1-(\frac{1}{5}{)}^2}$
$=-\frac{\sqrt{24}}{5}$
$=\frac{-2\sqrt{6}}{5}$