在△ABC中,角ABC所对边分别是abc且cosA=1/4 （1）求sin²B+C/2+cos2A的值 （2）若a=4,b+c=6,且b小于c求bc

在△ABC中,角ABC所对边分别是abc且cosA=1/4 （1）求sin²B+C/2+cos2A的值 （2）若a=4,b+c=6,且b小于c求bc
在△ABC中,角ABC所对边分别是abc且cosA=1/4 （1）求sin²B+C/2+cos2A的值 （2）若a=4,b+c=6,且b小于c求bc

在△ABC中,角ABC所对边分别是abc且cosA=1/4 （1）求sin²B+C/2+cos2A的值 （2）若a=4,b+c=6,且b小于c求bc
sin²(B+C)/2+cos2A
=1/2[1-cos(B+C)]+2cos^2A-1
=1/2(1+cosA)+2cos^2A-1
=1/2(1+1/4)+2*1/16-1
=-1/4
b+c=6 ①
根据余弦定理:
a^2=b^2+c^2-2bc*cosA
即16=b^2+c^2-2bc*1/4 ②
联合 ①②解方程组,且b小于c得
b=2,c=4

1
(sin(B+C)/2)^2=(cos(A/2))^2=(1+cosA)/2=5/8
cos2A=2(cosA)^2-1=-7/8
(sin(B+C)/2)^2+cos2A=-1/4
2
a=4
b+c=6
a^2=b^2+c^2-2bccosA=(b+c)^2-2bc(cosA+1)
bc=[(b+c)^2-a^2]/[2(cosA+1)]=20/[2(1/4+1)=40/5=8

求sin²B+C/2+cos2A的中的C/2是边还是角

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