三角形ABC所对的三边长为abc cos=1/4求sin方[（B+C）/2]+cos2A求a=4 ,b+c=6 且bcosA=1/4A+B+C=180(B+C)/2=180-A/21.sin^2[（B+C）/2]+cos2A=cos^2A/2+cos2A=(cosA+1)/2+2cos^2A-1=-1/42.余弦定理a^2=b^2+c^2-2bccosA...