cos(2A+B+C)=cos2A*cos(B+C)-sin2A*sin(A+B),
因A、B、C是三角形的内角,所以A+B+C=180
B+C=180-A代入前面的式中得cos(2A+B+C)=cos2A*cos(180-A)-sinA*sin(180-A)
=-cos2A*cocA-sin2A*sinA,
将倍角公式sin2A=2sinA*cosA 及Cos2A=Cos^2(a)-Sin^2(a) 代入得cos(2A+B+C)=-(Cos^2(A)-Sin^2(A) )*cosA- 2sinA*cosA *sinA=-Cos^2(A)*cosA+Sin^2(A) *cosA-2sin^2(A)*cosA
=-(Cos^3(A)+sin^2(A)*cosA )=-cosA*(Cos^2(A)+sin^2(A))=-cosA*1=-cosA
(注：Cos^2(A)+sin^2(A）=1),证明完毕