Question

If ${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=\pi$, show that $x^2+y^2+z^2+2xyz=1$

Answer 1

The given equation is ${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=\pi$ ......$\left(1\right)$

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

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

since ${\cos }^{-1}A+{\cos }^{-1}B={\cos }^{-1}[AB-\sqrt{1-A^2}\sqrt{1-B^2}]$

where $-1\le A,B\le 1$

$\therefore {\cos }^{-1}[xy-\sqrt{1-x^2}\sqrt{1-y^2}]=\pi -{\cos }^{-1}z$

$\Rightarrow \cos \left({\cos }^{-1}[xy-\sqrt{1-x^2}\sqrt{1-y^2}]\right)=\cos (\pi -{\cos }^{-1}z)$

$\Rightarrow xy-\sqrt{1-x^2}\sqrt{1-y^2}=\cos {\cos }^{-1}z$ since $\cos (\pi -\theta )=-\cos \theta$

$\Rightarrow xy-\sqrt{1-x^2}\sqrt{1-y^2}=-z$

$\Rightarrow xy+z=\sqrt{1-x^2}\sqrt{1-y^2}$

squaring both sides, we get

${(xy+z)}^2=(1-x^2)(1-y^2)$

$\Rightarrow x^2{y}^2+z^2+2xyz=1-x^2-y^2+x^2{y}^2$

$x^2+y^2+z^2+2xyz=1$

Hence proved.