Question

If $x,y,z\in [-1,1]$ such that ${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=0$, find $x+y+z$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

We have,

$x,y,z\in [-1,1]$

$-1\le x\le 1,-1\le y\le 1,-1\le z\le 1$

$0\le {\cos }^{-1}x\le \pi ,0\le {\cos }^{-1}y\le \pi ,0\le {\cos }^{-1}z\le \pi$

Therefore,

${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=0$

$⟹{\cos }^{-1}x=0,{\cos }^{-1}y=0,{\cos }^{-1}z=0$

$x=y=z=1$

Hence, $x+y+z=3$