在三角形abc中,角abc所对的边分别为abc又cosa=4/5,求(1)sin的平方bc/2+cos2a的值(

1.sin的平方(b+c)/2+cos2a
=(1-cos(b+c))/2+cos2a
=(1+cosa)/2+cos2a
=9/10+2cosa^2-1
=59/50

2. S=bcsina/2=3
csina=3

cosa=4/5, sina=3/5
c=5.

O(∩_∩)O哈哈~
1.sin的平方(b+c)/2+cos2a
=(1-cos(b+c))/2+cos2a
=(1+cosa)/2+cos2a
=9/10+2cosa^2-1
=59/50

2. S=bcsina/2=3
csina=3

cosa=4/5, sina=3/5
c=5.

解1.sin的平方(b+c)/2+cos2a
=(1-cos(b+c))/2+cos2a
=(1+cosa)/2+cos2a
=9/10+2cosa^2-1
=59/50

2. S=bcsina/2=3
csina=3

cosa=4/5, sina=3/5
c=5.