Question

三角形ABC所对的三边长为abc cos=1/4

Answer 1

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b

cosA=1/4

A+B+C=180

(B+C)/2=180-A/2

1.

sin^2[（B+C）/2]+cos2A=cos^2A/2+cos2A

=(cosA+1)/2+2cos^2A-1

=-1/4

2.

余弦定理

a^2=b^2+c^2-2bccosA

(b+c)^2=b^2+c^2+2bc

a^2=36-2bc-bc/2

bc=8