`cos alpha sin (beta-gamma) + cos beta sin (gamma-alpha) + cos gamma sin (alpha-beta)=`

If origin is the orthocentre of the triangle with vertices A(cos alpha,sin alpha),B(cos beta,sin beta),C(cos gamma,sin gamma) then cos(2 alpha-beta-gamma)+cos(2 beta-gamma-alpha)+cos(2 gamma-alpha-beta)=

Prove that : cos^2 (beta-gamma) + cos^2 (gamma-alpha) + cos^2 (alpha-beta) =1+2cos (beta-gamma) cos (gamma-alpha) cos (alpha-beta) .

sin (beta+ gamma- alpha) + sin (gamma+ alpha - beta) + sin (alpha + beta- gamma)- sin (alpha + beta + gamma)=

If Delta = det [[sin alpha, cos alpha, sin alpha + cos alphasin beta, cos alpha, sin beta + cos betasin gamma, cos alpha, sin gamma + cos beta]] then Delta

cos (alpha + beta) cos gamma-cos (beta + gamma) cos alpha = sin beta sin (gamma-alpha)

prove that | (sin alpha, sin beta, sin gamma), (cos alpha, cos beta, cos gamma), (sin 2alpha, sin 2beta, sin 2gamma) | = 4sin (1/2) (beta-gamma) * sin (1/2) (gamma-alpha) * sin (1/2) (alpha-beta) xx {sin (beta + gamma) + sin (gamma + alpha) + sin (alpha + beta)}.