If `sinalpha+sinbeta+singamma=0=cosalpha+cosbeta+cosgamma` then `sum cos (alpha-beta)=`

If sinalpha+sinbeta+singamma=cosalpha+cosbeta+cosgamma=0 then prove that cos3alpha+cos3beta+cos3gamma=3cos(alpha+beta+gamma)

If sinalpha + sinbeta + singamma = 3 , then cosalpha + cosbeta + cosgamma equals :

cos(beta-gamma)+cos(gamma-alpha)+cos(alpha-beta)=-3/2 show that cosalpha+cosbeta+cosgamma=0 and sinalpha+sinbeta+singamma=0 also cos(beta-gamma)=cos(gamma-alpha)=cos(alpha-beta)=-1/2 .

If cosalpha+cosbeta+cosgamma=0=sinalpha+sinbeta+singamma then cos(2alpha-beta-gamma)+cos(2beta-gamma-alpha)+cos(2gamma-alpha-beta)=

If (cosalpha+cosbeta+cosgamma)=0, sinalpha+sinbeta+singamma=0 then show that cos3alpha+cos3beta+cos3gamma=3cos(alpha+beta+gamma)

Let A and B denote the statement : A : cosalpha + cosbeta + cosgamma = 0 , B : sinalpha + sinbeta + singamma = 0 If cos(beta - gamma) + (gamma - alpha) + cos (alpha - beta) =-3/2 , then :

If sinalpha-sinbeta=a and cosalpha+cosbeta=b then write the value of "cos"(alpha+beta) .

If cosalpha+coabeta+cosgamma=0=sinalpha+sinbeta+singamma then cos(alpha+beta)+cos(beta+gamma)+cos(gamma+alpha)=