Question

The value of $\cos ({\cos }^{-1}\left(\frac{-1}{\sqrt{2}}\right)+\frac{\pi }{4})$ is

$\left(a\right)0\left(b\right)1$
$\left(c\right)\frac{1}{2}\left(d\right)-1$

Answer 1

We know that, ${\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$
$\therefore \cos ({\cos }^{-1}\left(\frac{-1}{\sqrt{2}}\right)+\frac{\pi }{4})=\cos (\pi -{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)+\frac{\pi }{4})$
$=\cos (\pi -\frac{\pi }{4}+\frac{\pi }{4})$
$=\cos \pi$
$=-1$
Ans : $\left(d\right)$