三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2]
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![三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2]](/uploads/image/z/974170-10-0.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2ABC%2C%E6%B1%82%E8%AF%81cos%28A%2BB%29%3D-cosC%2Ccos%5B%28A%2BB%29%2F2%5D%3Dsin%28C%2F2%29%E5%92%8Csin%283A%2B3B%29%3Dsin%283C%29%2Csin%5B%283A%2B3B%29%2F2%5D%3D-cos%5B%283C%29%2F2%5D)
三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2]
三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2]
三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2]
A+B=π-C
cos(A+B)=cos(π-C)=-cosC
cos[(A+B)/2]=cos[(π-C)/2]=cos(π/2-C/2)=sin(C/2)
sin(3A+3B)=sin3*(A+B)=sin3*(π-C)=sin(3π-3C)=sin(3C)
sin[(3A+3B)/2]=-cos[(3C)/2]
sin[(3A+3B)/2]=sin[3/2*(A+B)]=sin[3/2*(π-C)]=sin(3/2π-3/2*C)=sin(3C)
A+B=180°-C
cos(A+B)=cos(180°-C)=-cosC
cos[(A+B)/2]=cos[(180°-C)/2]=cos(180°/2-C/2)=sin(C/2)
sin(3A+3B)=sin3*(A+B)=sin3*(180°-C)=sin(3*180°-3C)=sin(3C)
sin[(3A+3B)/2]=-cos[(3C)/2]
sin[(3A+3B)/2]=sin[3/2*(A+B)]=sin[3/2*(180°-C)]=sin(270°-3*C/2)=cos[(3C)/2]