Question

在三角形ABC中,角A,B,C,的对边分别为a,b,c 且COS(AB,AC)=1/4 求sin^2B+C/2+COS2A的值

Answer 1

cosA=1/4
[sin(B+C)/2]^2=[1-cos(B+C)]/2=(1+cosA)/2=5/8
cos2A=2cosA^2-1=-7/8
原式＝5/8-7/8=-1/4

Answer 2

sin（π/2－A）＝cosA
cos(2α)=(cosα)^2-(sinα)^2=2(cosα)^2-1=1-2(sinα)^2
COSA=2(COS(A/2))^2-1=1/4
COS^2(A/2)=(1+1/4)/2=5/8
sin^2((B+C)/2)+COS2A=sin^2((π-A)/2)+COS2A=COS^2(A/2)+2(COSA)^2-1
=COS^2(A/2)-7/8=-1/4