Question

If $\cos (y-z)+\cos (z-x)+\cos (x-y)=-\frac{3}{2}$, prove that $\cos x\cos y\cos z=0$ = $\sin x$ +$\sin y+\sin z$

Answer 1

$cos(z-x)+cos(y-2)+cos(x-y)=\frac{-3}{2}$

$cosxcosy+sinxsiny+cosycos3+sinysin2+coszcosx+sin2sinx=\frac{-3}{2}...\left(1\right)$

$\Rightarrow 3+2(cosxcosy+....)+2(sinxsiny....)=0...\left(2\right)$

3 can be written as 1 + 1 + 1 s.t

$(co{s}^{2}x+si{n}^{2x})+(co{s}^{2}y+si{n}^{2}y)+(co{s}^2+si{n}^{2}2)$

so, the equation becomes,

$0=co{s}^{2}x+co{s}^{2}y+co{s}^{2}z+2cosxcosy+2cosycosz+$

$2coszcosx+si{n}^{2}x+si{n}^{2}y+si{n}^{2}z+2sinxsiny$

$+2sinysinz+2sinzsinx$

$0=(cosx+cosy+cosz{)}^2+(sinx+siny+sinz{)}^2$

This is possible if,

$cosx+cosy+cosz=0$

or $sinx+siny+sinz=0$

$\therefore cosx+cosy+cosz=0=sinx+siny+sinz$