$x=cosalpha+isinalpha$ , $y=cosbeta+isinbeta$ , $z=cosgamma+isingamma$ and $x+y+z=xyz$ . prove that $cos(beta-gamma)+cos(gamma-alpha)+cos(alpha-beta)=1$ ?

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Answer 1 · 2016-11-05T13:22:06

This proposition is false, e.g. when $alpha = 0$ , $beta = pi/2$ , $gamma=-pi/2$

Let:

${ (alpha = 0), (beta = pi/2), (gamma = -pi/2) :}$

Then:

${ (x = 1), (y = i), (z = -i) :}$

and:

$x + y + z = 1+i-i = 1 = 1*i*(-i) = xyz$

With these values, we find:

$cos(beta-gamma)+cos(gamma-alpha)+cos(alpha-beta)$

$=cos(pi)+cos(-pi/2)+cos(-pi/2)$

$= -1+0+0=-1 != 1$

x=cosalpha+isinalphax=cosalpha+isinalpha,y=cosbeta+isinbetay=cosbeta+isinbeta,z=cosgamma+isingammaz=cosgamma+isingamma andx+y+z=xyzx+y+z=xyz. prove that cos(beta-gamma)+cos(gamma-alpha)+cos(alpha-beta)=1cos(beta-gamma)+cos(gamma-alpha)+cos(alpha-beta)=1?