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cos(2A) + cos(2B) + cos(2C) + cos(2A)cos(2B)cos(2C)
　= (1/2){1+cos(2A)}{1+cos(2B)}{1+cos(2C)} - (1/2){1-cos(2A)}{1-cos(2B)}{1-cos(2C)}
　= 4{cos(A)cos(B)cos(C)}^2 - 4{sin(A)sin(B)sin(C)}^2,　　　　　　････ (1)

cos(2A)cos(2B)cos(2C) - cos(2A+2B+2C) = cos(2A)sin(2B)sin(2C) + sin(2A)cos(2B)sin(2C) + sin(2A)sin(2B)cos(2C)
　= {cos(A)^2 -sin(A)^2}sin(2B)sin(2C) + sin(2A){cos(B)^2 -sin(B)^2}sin(2C) + sin(2A)sin(2B){cos(C)^2 -sin(C)^2}
　= 4cos(A)cos(B)cos(C){cos(A)sin(B)sin(C) +sin(A)cos(B)sin(C) +sin(A)sin(B)cos(C)}
　 -4sin(A)sin(B)sin(C){sin(A)cos(B)cos(C) +cos(A)sin(B)cos(C) +cos(A)cos(B)sin(C)}
　= 4cos(A)cos(B)cos(C){cos(A)cos(B)cos(C) -cos(A+B+C)}
　 -4sin(A)sin(B)sin(C){sin(A)sin(B)sin(C) +sin(A+B+C)}
　= [ 1 ] -4cos(A)cos(B)cos(C)cos(A+B+C) -4sin(A)sin(B)sin(C)sin(A+B+C), ･････(2)

(1)-(2) より
　cos(2A) + cos(2B) + cos(2C) + cos(2A+2B+2C) = 4cos(A)cos(B)cos(C)cos(A+B+C) + 4sin(A)sin(B)sin(C)sin(A+B+C),

A+B+C+D=0 のとき
　cos(2A) + cos(2B) + cos(2C) + cos(2D) = 4cos(A)cos(B)cos(C)cos(D) - 4sin(A)sin(B)sin(C)sin(D),